3 無限小解析

3.2

問2

(i)

部分積分により
$$ \int_{a}^{x} t^{\alpha}\log{t}\,\d{t} = \left[\frac{1}{\alpha+1}t^{\alpha+1}\log{t}\right]_{a}^{x} - \frac{1}{\alpha+1} \int_{a}^{x} t^{\alpha} \,\d{t}$$
となる．$x \to +\infty$のとき$t^{\alpha} = o(t^{\alpha}\log{t})$であって
$$ \int_{a}^{+\infty} t^{\alpha}\log{t} \,\d{t} = \lim_{x\to+\infty}\frac{x^{\alpha+1}}{\alpha+1}\left(\log{x}-\frac{1}{\alpha+1}\right) + \text{const.} = +\infty$$
である．よって$\int_{a}^{x} t^{\alpha}\log{t}\,\d{t}$は無限大であって
$$ \int_{a}^{x} t^{\alpha} \,\d{t} = o\left(\int_{a}^{x} t^{\alpha}\log{t} \,\d{t} \right) \quad(x \to +\infty)$$
が成り立つので，
$$ \int_{a}^{x} t^{\alpha}\log{t} \,\d{t} - \frac{x^{\alpha+1}}{\alpha+1}\log{x} = o\left(\int_{a}^{x} t^{\alpha}\log{t} \,\d{t} \right) \quad(x\to+\infty) \quad\leadsto\quad \int_{a}^{x} t^{\alpha}\log{t} \,\d{t} \sim \frac{x^{\alpha+1}}{\alpha+1} \log{x}$$
を得る．

(ii)

部分積分により
$$ \int_{a}^{x} t(\log{t})^{\alpha} \,\d{t} = \left[\frac{t^{2}}{2}(\log{t})^{\alpha}\right]_{a}^{x} - \frac{\alpha}{2} \int_{a}^{x} t(\log{t})^{\alpha-1} \,\d{t}$$
となる．$x\to+\infty$のとき$t(\log{t})^{\alpha-1} = o(t(\log{t})^{\alpha})$である．また，$t > \max\{1,e^{\frac{\alpha}{2}}\} \eqqcolon a'$のとき
$$ \frac{1}{t} < t(\log{t})^{\alpha}$$
となるので，
$$ +\infty = \int_{a'}^{+\infty} \frac{1}{t} \,\d{t} \leq \int_{a'}^{\infty} t(\log{t})^{\alpha} \,\d{t}$$
である．よって$\int_{a}^{x} t(\log{t})^{\alpha} \,\d{t}$は無限大であって
$$ \int_{a}^{x} t(\log{t})^{\alpha-1} \,\d{t} = o\left(\int_{a}^{x} t(\log{t})^{\alpha} \,\d{t} \right) \quad(x\to+\infty)$$
が成り立つので，
$$ \int_{a}^{x} t(\log{t})^{\alpha} \,\d{t} - \frac{x^{2}}{2}(\log{x})^{\alpha} = o\left(\int_{a}^{x} t(\log{t})^{\alpha} \,\d{t}\right)\quad(x\to+\infty) \quad\leadsto\quad \int_{a}^{x} t(\log{t})^{\alpha} \,\d{t} \sim \frac{x^{2}}{2}(\log{x})^{\alpha}$$
を得る．

3.3

問2

(i)

$$ \sin^{2}x = \frac{1}{2}(1-\cos{2x}) = x^{2}-\frac{x^{4}}{3}+o(x^{4})$$
より
$$ \frac{1}{\sin^{2}x} = \frac{1}{x^{2}\left(1-\frac{x^{2}}{3}+o(x^{2})\right)} = \frac{1}{x^{2}}\left(1+\frac{x^{2}}{3} +o(x^{2})\right) = \frac{1}{x^{2}}+\frac{1}{3}+o(1)$$
となるので，
$$ \lim_{x\to 0} \left(\frac{1}{x^{2}} - \frac{1}{\sin^{2}x}\right) = \lim_{x\to 0} \left(-\frac{1}{3} + o(1)\right) = -\frac{1}{3}$$
となる．

(ii)

$$ \frac{x-\sin{x}}{x(1-\cos{x})} = \frac{\frac{x^{3}}{6}+o(x^{3})}{\frac{x^{3}}{2}+o(x^{3})} = \frac{\frac{1}{6}+o(1)}{\frac{1}{2}+o(1)} \to \frac{1}{3}.$$

(iii)

$$ \cot{x} = \frac{\cos{x}}{\sin{x}} = \frac{1-\frac{x^{2}}{2}+o(x^{2})}{x-\frac{x^{3}}{6}+o(x^{3})} = \frac{1}{x}\left(1-\frac{x^{2}}{2}+o(x^{2})\right)\left(1+\frac{x^{2}}{6}+o(x^{2})\right) = \frac{1}{x}\left(1-\frac{x^{2}}{3}-\frac{x^{4}}{12}+o(x^{4})\right)$$
より
$$ \lim_{x\to 0} \left(\frac{1}{x}-\cot{x}\right) = \lim_{x\to 0} \left(\frac{x}{3}+\frac{x^{3}}{12}+o(x^{3})\right) = 0$$
となる．

3.4

問1

(i)

$\sin{x}=x+o(x^{2}) = x(1+o(x))$より
$$ x\log\sin{x} = x(\log{x}+\log(1+o(x))) = x\log{x}+x\log(1+o(x)) = x\log{x}+o(x\log{x}) \eqqcolon u$$
となるので，
\begin{align} (\sin{x})^{x} &= \exp(u) \\ &= 1 + u + \frac{u^{2}}{2} + o(u^{2}) \\ &= 1+x\log{x}+\frac{1}{2}(x\log{x})^{2} + o((x\log{x})^{2}) \quad(x\to +0) \end{align}
を得る．

(ii)

$$ \log(1+x^{-\frac{1}{2}}) = x^{-\frac{1}{2}}-\frac{1}{2}x^{-1} + \frac{1}{3}x^{-\frac{3}{2}} + o(x^{-\frac{3}{2}}) = x^{-\frac{1}{2}}\left(1-\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{3}x^{-1}+o(x^{-1})\right)$$
であるから，
$$ u \coloneqq -\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{3}x^{-1}+o(x^{-1})$$
とおけば，
\begin{align} \sqrt{\log\left(1+\frac{1}{\sqrt{x}}\right)} &= x^{-\frac{1}{4}}(1+u)^{\frac{1}{2}} \\ &= x^{-\frac{1}{4}}\left(1+\frac{1}{2}u-\frac{1}{8}u^{2}+o(u^{2})\right) \\ &= x^{-\frac{1}{4}}\left(1-\frac{1}{4}x^{-\frac{1}{2}} + \left(\frac{1}{6}-\frac{1}{32}\right)x^{-1}+o(x^{-1})\right) \\ &= x^{-\frac{1}{4}} - \frac{1}{4}x^{-\frac{3}{4}} + \frac{13}{96}x^{-\frac{5}{4}}+o(x^{-\frac{5}{4}}) \quad(x\to+\infty) \end{align}
を得る．

問2

(i)

$$ \log(x-1) = \log{x} + \log(1-x^{-1}) = \log{x} + (-x^{-1}+o(x^{-1})) = \log{x} - \frac{1}{x} + o\left(\frac{1}{x}\right) \quad(x\to+\infty).$$

(ii)

$$ \log(x+1) = \log{x}+\log(1+x^{-1}) = \log{x}+\left(\frac{1}{x}-\frac{1}{2x^{2}}+o\left(\frac{1}{x^{2}}\right)\right) = \log{x}\left(1+\frac{1}{x\log{x}}-\frac{1}{2x^{2}\log{x}}+o\left(\frac{1}{x^{2}\log{x}}\right)\right)$$
であるから，
$$ u \coloneqq \frac{1}{x\log{x}}-\frac{1}{2x^{2}\log{x}}+o\left(\frac{1}{x^{2}\log{x}}\right)$$
とおけば，
\begin{align} \sqrt{\log(x+1)} &= \sqrt{\log{x}}(1+u)^{\frac{1}{2}}\\ &= \sqrt{\log{x}}\left(1+\frac{1}{2}u-\frac{1}{8}u^{2}+o(u^{2})\right) \\ &= \sqrt{\log{x}}\left(1+\frac{1}{2x\log{x}}-\frac{1}{4x^{2}\log{x}}-\frac{1}{8(x\log{x})^{2}}+o\left(\frac{1}{(x\log{x})^{2}}\right)\right)\\ &= \sqrt{\log{x}} + \frac{1}{2x\sqrt{\log{x}}}-\frac{1}{4x^{2}\sqrt{\log{x}}} - \frac{1}{8x^{2}\log{x}\sqrt{\log{x}}} +o\left(\frac{1}{x^{2}\log{x}\sqrt{\log{x}}}\right) \\ &= \sqrt{\log{x}} + \frac{1}{2x\sqrt{\log{x}}}-\frac{1}{4x^{2}\sqrt{\log{x}}} +o\left(\frac{1}{x^{2}\sqrt{\log{x}}}\right) \quad(x\to+\infty) \end{align}
を得る．

3.7

問1

(i)

$$ \frac{1}{\log{n}} \geq \frac{1}{n},\ n \geq 2 \quad\leadsto\quad \sum \frac{1}{\log{n}} \geq \sum \frac{1}{n}=+\infty.$$

(ii)

$$ n^{-\frac{a}{n}} \geq n^{-1},\ n \geq a \quad\leadsto\quad \sum n^{-\frac{a}{n}} \geq \sum n^{-1} = +\infty.$$

(iii)

$$ a_{n} \coloneqq \frac{\sqrt{n}}{1+n^{2}} \sim \frac{1}{n^{\frac{3}{2}}} \quad(n\to\infty).$$

(iv)

$$ \frac{1}{n^{\log{n}}} \leq \frac{1}{n^{2}},\ n \geq e^{2}.$$

(v)

$$ \sqrt[n]{\frac{1}{(\log{n})^{n}}} = \frac{1}{\log{n}} \to 0<1 \quad(n\to\infty).$$

(vi)

$$ a^{\log{n}} = \exp(\log{n}\log{a}) = n^{\log{a}} = \frac{1}{n^{-\log{a}}}.$$

(vii)

$$ \frac{1}{(\log{n})^{\log{n}}} = \exp(-\log{n}\log{n}) \leq \exp(-2\log{n}) = \frac{1}{n^{2}},\ n \geq e^{2}.$$

(viii)

$$ \frac{(n!)^{2}}{(2n)!} \cdot \frac{(2(n-1))!}{((n-1)!)^{2}} = \frac{n^{2}}{2n(2n-1)} \to \frac{1}{4} < 1 \quad(n\to\infty).$$

(ix)

$a=1$のときは明らかに収束する．$a\neq 1$のとき，$a=1\pm h,\,h>0,\,$とおくと
$$ a^{\frac{1}{n}}-1 = \pm\frac{h}{n} + o\left(\frac{1}{n}\right) \sim \pm\frac{h}{n} \quad(n\to\infty)$$
となるので，$\sum (a^{\frac{1}{n}}-1)$は発散する．

(x)

$$ n\log \left(1-\frac{1}{n}\log{n}\right) = n \left(-\frac{\log{n}}{n} + o\left(\frac{1}{n}\right)\right) = -\log{n} +o(1) \quad(n\to\infty)$$
より
$$ \left(1-\frac{1}{n}\log{n}\right)^{n} = \frac{1}{n}e^{o(1)} \sim \frac{1}{n} \quad(n\to\infty)$$
となるので，$\sum (1-(\log{n})/n)^{n}$は発散する．

(xi)

$$ n^{\frac{1}{n}} = \exp \left(\frac{\log{n}}{n}\right) = 1 + \frac{\log{n}}{n} + o\left(\frac{\log{n}}{n}\right) \quad(n\to\infty)$$
より
$$ \frac{1}{n^{p}(n^{\frac{1}{n}}-1)} = \frac{1}{n^{p-1}\log{n}}(1+o(1))^{-1} \sim \frac{1}{n^{p-1}\log{n}} \quad(n\to\infty)$$
であり，
$$ \sum \frac{1}{n^{p-1}\log{n}}\ \begin{dcases} \leq \sum \frac{1}{n^{p-1}} < +\infty & p-1 > 1 \\ \geq \sum \frac{1}{n\log{n}} = +\infty & p-1 \leq 1 \end{dcases}$$
であるから，$\sum (n^{p}(n^{1/n}-1))^{-1}$は$p>2$のとき収束し$p\leq 2$のとき発散する．

(xii)

\begin{align} n^{2}\log\frac{n}{n+1} = n^{2}\log \left(1-\frac{1}{n+1}\right) &= n^{2}\left(-\frac{1}{n+1}-\frac{1}{2(n+1)^{2}} + o\left(\frac{1}{(n+1)^{2}}\right)\right) \\ &= n \frac{-n}{n+1} - \frac{1}{2}\left(\frac{n}{n+1}\right)^{2} + o(1) \\ &= n\left(-1+\frac{1}{n+1}\right) - \frac{1}{2}\left(1-\frac{1}{n+1}\right)^{2} + o(1) \\ &= -n + \left(1-\frac{1}{n+1}\right) -\frac{1}{2}\left(1-\frac{1}{n+1}\right)^{2} + o(1) \\ &= -n + \frac{1}{2}-\frac{1}{2}\frac{1}{(n+1)^{2}} + o(1) \\ &= -n +\frac{1}{2} +o(1) \quad(n\to\infty) \end{align}
より
$$ \left(\frac{n}{n+1}\right)^{n^{2}} \sim \sqrt{e}\cdot e^{-n} \quad(n\to\infty)$$
となるので，$\sum (n/(n+1))^{n^{2}}$は収束する．

(xiii)

$$ \frac{n^{q}}{(n+1)^{p+q}} = \left(\frac{n}{n+1}\right)^{q} \frac{1}{(n+1)^{p}} \sim \frac{1}{n^{p}} \quad(n\to\infty).$$

(xiv)

$$ \frac{a^{n}}{n!}n^{\frac{n}{2}} \cdot \frac{(n-1)!}{a^{n-1}}\frac{1}{(n-1)^{\frac{n-1}{2}}} = \frac{a}{n}\left(\frac{n}{n-1}\right)^{\frac{n}{2}}\sqrt{n-1} = a\left(1-\frac{1}{n}\right)^{-\frac{n}{2}}\frac{\sqrt{n-1}}{n} \sim a \frac{e^{\frac{1}{2}}}{\sqrt{n}} \to 0 \quad(n\to\infty).$$

(xv)

$$ \frac{1}{n\left(1+\frac{1}{2} +\cdots+ \frac{1}{n}\right)} \sim \frac{1}{n\log{n}} \quad(n\to\infty).$$

(xvi)

(xii)での計算より
$$ a_{n} \coloneqq \left(\frac{n}{n+1}\right)^{n^{2}}e^{n} \sim \sqrt{e} \not\to 0 \quad(n\to\infty)$$
となるので，$\sum a_{n}$は発散する．

(xvii)

$$ \left(\frac{(2n-1)!!}{(2n)!!}\right)^{p} \cdot \left(\frac{(2(n-1))!!}{(2(n-1)-1)!!}\right)^{p} = \left(\frac{2n-1}{2n}\right)^{p} = \left(1-\frac{1}{2n}\right)^{p} = 1-\frac{\frac{p}{2}}{n} + O\left(\frac{1}{n^{2}}\right) \quad(n\to\infty)$$
となるので，ガウスの判定法により，$\sum ((2n-1)!!/(2n)!!)^{p}$は$p>2$のとき収束し$p \leq 2$のとき発散する．

(xviii)

$$ \frac{(n!)^{2}}{\prod_{k=1}^{n}(k^{2}+k+3)} \cdot \frac{\prod_{k=1}^{n-1} (k^{2}+k+3)}{((n-1)!)^{2}} = \frac{n^{2}}{n^{2}+n+3} = \frac{1}{1+\frac{1}{n}+\frac{3}{n^{2}}} = 1-\frac{1}{n}+O\left(\frac{1}{n^{2}}\right) \quad(n\to\infty).$$

3.10

問1

(i)

$$ \frac{\log(n^{2}+1)}{n} = 2\frac{\log{n}}{n} + \frac{1}{n}\log\left(1+\frac{1}{n^{2}}\right) = 2 \frac{\log{n}}{n} + \frac{1}{n}\left(\frac{1}{n^{2}} + o\left(\frac{1}{n^{2}}\right)\right) = 2\frac{\log{n}}{n} + o\left(\frac{\log{n}}{n} \right) \quad(n\to\infty)$$
となるので，$\prod \sqrt[n]{n^{2}+1}$は発散する（cf. 命題3.34）．

(ii)

$$ \log{a^{\frac{1}{n^{p}}}} = \frac{\log{a}}{n^{p}}.$$

(iii)

$p>1$のとき，$p>p'>1$を取ると
$$ \frac{\log{n}}{n^{p}} \leq \frac{1}{(p-p')n^{p'}} \quad\leadsto\quad \sum \frac{\log{n}}{n^{p}} < +\infty$$
となるので，$\prod n^{\frac{1}{n^{p}}}$は収束する．$p \leq 1$のとき，
$$ \frac{\log{n}}{n^{p}} \geq \frac{1}{n^{p}},\ n \geq e \quad\leadsto\quad \sum \frac{\log{n}}{n^{p}} = +\infty$$
となるので，$\prod n^{\frac{1}{n^{p}}}$は発散する．

(iv)

$$ \frac{2(1+2n+3n^{2})}{4+5n+6n^{2}} = 1-\frac{2+n}{4+5n+6n^{2}}$$
であり
$$ \frac{2+n}{4+5n+6n^{2}} \sim \frac{1}{6n} \quad(n\to\infty)$$
より$\sum (2+n)/(4+5n+6n^{2})$は発散するので，$\prod 2(1+2n+3n^{2})/(4+5n+6n^{2})$も発散する（cf. 定理3.39）．

(v)

$$ \frac{n^{2}-\sqrt{n}+3}{n^{2}+1} = 1- \frac{\sqrt{n}-2}{n^{2}+1},\ \frac{\sqrt{n}-2}{n^{2}+1} \sim \frac{1}{n^{\frac{3}{2}}} \quad(n\to\infty).$$

(vi)

$$ \frac{5n^{3}+(-1)n^{2}+1}{5n^{3}+2} = 1 + \frac{(-1)^{n}n^{2}-1}{5n^{3}+2} \eqqcolon 1 +u_{n}$$
とおくと，
$$ u_{n}^{2} \sim \frac{1}{25n^{2}} \quad(n\to\infty)$$
より$\sum u_{n}^{2}$は収束し，
$$ u_{n} \sim \frac{(-1)^{n}}{5n} \quad(n\to\infty)$$
より$\sum u_{n}$も収束するので，$\prod (1+u_{n})$は収束する（cf. 定理3.36）．

(vii)

$$ \frac{2\sum_{k=1}^{n} k}{n^{2}} = \frac{n(n+1)}{n^{2}} = 1+\frac{1}{n}.$$

(viii)

$$ \sqrt{n(n+1)} = n\left(1+\frac{1}{n}\right)^{\frac{1}{2}} = n\left(1+\frac{1}{2n} - \frac{1}{8n^{2}} +o\left(\frac{1}{n^{2}}\right)\right) = n+\frac{1}{2}-\frac{1}{8n}+o\left(\frac{1}{n}\right) \quad(n\to\infty)$$
より
$$ 2\left(\sqrt{n(n+1)}-n\right) = 1-\frac{1}{4n}+o\left(\frac{1}{n}\right) \quad(n\to\infty)$$
となるので，$\prod 2(\sqrt{n(n+1)}-n)$は発散する．

(ix)

$$ n\sin\frac{1}{n} = n\left(\frac{1}{n}-\frac{1}{6n^{3}} +o\left(\frac{1}{n^{3}}\right)\right) = 1 -\frac{1}{6n^{2}} +o\left(\frac{1}{n^{2}}\right) \quad(n\to\infty).$$

(x)

$$ \tan\frac{1}{n} = \frac{\frac{1}{n}-\frac{1}{6n^{3}}+o\left(\frac{1}{n^{3}}\right)}{1-\frac{1}{2n^{2}}+o\left(\frac{1}{n^{2}}\right)} = \left(\frac{1}{n}-\frac{1}{6n^{3}}+o\left(\frac{1}{n^{3}}\right)\right) \left(1+\frac{1}{2n^{2}}+o\left(\frac{1}{n^{2}}\right)\right) = \frac{1}{n} +\frac{1}{3n^{3}} + o\left(\frac{1}{n^{3}}\right) \quad(n\to\infty)$$
より
$$ n\tan\frac{1}{n} = 1+\frac{1}{3n^{2}} +o\left(\frac{1}{n^{2}}\right) \quad(n\to\infty)$$
となるので，$\prod n\tan{\frac{1}{n}}$は収束する．

演習問題

1

(i)

$$ \log\frac{x-a}{x+a} = \log \left(1-\frac{2a}{x+a}\right) = -\frac{2a}{x+a} + o\left(\frac{1}{x+a}\right) \quad(x\to+\infty).$$

(ii)

$$ \cos{x} = 1-\frac{x^{2}}{2} +o(x^{2}) \quad(x\to 0) \quad\leadsto\quad \log\cos{x} = -\frac{x^{2}}{2}+o(x^{2}) \quad(x \to 0).$$

(v)

$$ e^{\frac{1}{x}}-1 = \frac{1}{x} +o\left(\frac{1}{x}\right) \quad(x\to+\infty).$$

(vii)

$$ \cos{x}=1-\frac{x^{2}}{2}+o(x^{2}) \quad(x\to 0) \quad\leadsto\quad \cos(1-\cos{x}) = 1-\frac{1}{8}x^{4} +o(x^{4}) \quad(x\to 0).$$

(ix)

$$ \frac{\frac{1}{(x+2)^{2}+(x+1)^{2}}}{-\frac{1}{x^{2}}} = -\frac{x^{2}}{2x^{2}+6x+5} \to -\frac{1}{2} \quad(x\to+\infty) \quad\leadsto\quad \lim_{x\to+\infty} x\left(\arctan\frac{x+1}{x+2}-\frac{\pi}{4}\right) = \lim_{x\to+\infty} \frac{\arctan\frac{x+1}{x+2}-\frac{\pi}{4}}{\frac{1}{x}} = -\frac{1}{2}.$$

(x)

$$ \frac{\sin{x}}{x} = 1-\frac{x^{2}}{6} +o(x^{2}) \quad(x\to 0) \quad\leadsto\quad \log\frac{\sin{x}}{x} = -\frac{x^{2}}{6}+o(x^{2}) \quad(x\to 0)$$
より
$$ \frac{\sin{x}}{x-\sin{x}}\log\frac{\sin{x}}{x} = \frac{-\frac{x^{3}}{6}+o(x^{3})}{\frac{x^{3}}{6}+o(x^{3})} \to -1 \quad(x\to 0)$$
となるので，
$$ \lim_{x\to 0} \left(\frac{\sin{x}}{x}\right)^{\frac{\sin{x}}{x-\sin{x}}} = e^{-1}$$
を得る．

2

(i)

$$ 0 < e^{-t^{2}} \to 0 \quad(t\to+\infty),\ \int_{0}^{+\infty} e^{-t^{2}} \,\d{t} = \frac{\sqrt{\pi}}{2}$$
より$\int_{x}^{+\infty} e^{-t^{2}}\,\d{t}$は$x\to+\infty$のとき無限小である．いま，部分積分により
$$ \int_{x}^{+\infty} e^{-t^{2}}\,\d{t} = \left[\frac{1}{-2t}e^{-t^{2}}\right]_{x}^{+\infty} - \frac{1}{2}\int_{x}^{+\infty} \frac{e^{-t^{2}}}{t^{2}} \,\d{t} = \frac{e^{-x^{2}}}{2x} - \frac{1}{2}\int_{x}^{+\infty} \frac{e^{-t^{2}}}{t^{2}} \,\d{t}$$
であり，
$$ \frac{e^{-t^{2}}}{t^{2}} = o(e^{-t^{2}}) \quad(t\to+\infty) \quad\leadsto\quad \int_{x}^{+\infty} \frac{e^{-t^{2}}}{t^{2}}\,\d{t} = o\left(\int_{x}^{+\infty} e^{-t^{2}} \,\d{t}\right)$$
であるから（cf. 定理3.7），
$$ \int_{x}^{+\infty} e^{-t^{2}} \,\d{t} \sim \frac{e^{-x^{2}}}{2x} \quad(x\to+\infty)$$
が成り立つ．

(ii)

$$ \log(1+t^{2}) = 2\log{t} + \log\left(1+\frac{1}{t^{2}}\right)$$
より
$$ \int_{0}^{x} \log(1+t^{2}) \,\d{t} = 2[t\log{t}-t]_{0}^{x} +\int_{0}^{x} \log \left(1+\frac{1}{t^{2}}\right) \,\d{t} = 2x\log{x} - 2x + \int_{0}^{x} \log \left(1+\frac{1}{t^{2}} \right) \,\d{t}$$
となる．ところで，
$$ \log \left(1+ \frac{1}{t^{2}} \right) = o(1) \quad(t\to+\infty)$$
より
$$ \int_{0}^{x} \log \left(1+\frac{1}{t^{2}} \right) \,\d{t} = o\left(\int_{0}^{x} 1 \,\d{t} \right) = o(x) \quad(x\to+\infty)$$
であるから（cf. 定理3.7），
$$ \int_{0}^{x} \log(1+t^{2})\,\d{t} = 2x\log{x} - 2x + o(x) \quad(x\to+\infty)$$
が成り立つ．

3

(i)

\begin{align} \tan{x} = \frac{\sin{x}}{\cos{x}} &= \frac{x-\dfrac{x^{3}}{6}+\dfrac{x^{5}}{120} + o(x^{5})}{1-\dfrac{x^{2}}{2} + \dfrac{x^{4}}{24} + o(x^{4})} \\ &= x \frac{1-\dfrac{x^{2}}{6}+\dfrac{x^{4}}{120} + o(x^{4})}{1-\dfrac{x^{2}}{2} + \dfrac{x^{4}}{24} + o(x^{4})} \\ &= x \left(1-\frac{x^{2}}{6}+\frac{x^{4}}{120} + o(x^{4})\right)\left(1+\frac{x^{2}}{2} +\left(\frac{1}{4}-\frac{1}{24}\right)x^{4} +o(x^{4})\right) \\ &= x \left(1 + \frac{x^{2}}{3} + \left(\frac{5}{24}+\frac{1}{120}-\frac{1}{12}\right) x^{4} +o(x^{4})\right) \\ &= x + \frac{1}{3}x^{3} +\frac{2}{15}x^{5} +o(x^{5}) \quad(x\to 0) \\ &\eqqcolon u \end{align}
より
\begin{align} \log(1+\tan{x}) = \log(1+u) &= u-\frac{u^{2}}{2}+\frac{u^{3}}{3}-\frac{u^{4}}{4} +O(u^{5}) \\ &= x-\frac{1}{2}x^{2} + \left(\frac{1}{3}+\frac{1}{3}\right)x^{3} + \left(-\frac{1}{2}\frac{2}{3} - \frac{1}{4}\right)x^{4} +O(x^{5}) \\ &= x-\frac{1}{2}x^{2} + \frac{2}{3}x^{3} - \frac{7}{12}x^{4} +O(x^{5}) \quad(x\to 0) \end{align}
を得る．

(ii)

$$ \cos{x} = 1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+O(x^{8}) \eqqcolon 1+u \quad(x\to 0)$$
より
\begin{align} \log\cos{x} = \log(1+u) &= u-\frac{u^{2}}{2}+\frac{u^{3}}{3}+O(u^{4}) \\ &= -\frac{1}{2}x^{2}+\left(\frac{1}{24}-\frac{1}{2}\frac{1}{4}\right)x^{4}+\left(-\frac{1}{720}-\frac{1}{2}\left(-\frac{1}{24}\right)+\frac{1}{3}\left(-\frac{1}{8}\right)\right)x^{6} +O(x^{8}) \\ &= -\frac{1}{2}x^{2}-\frac{1}{12}x^{4}-\frac{1}{45}x^{6}+O(x^{8}) \quad(x\to 0) \end{align}
を得る．

(iii)

$$ x\sin{x} = x^{2}-\frac{x^{4}}{6} + \frac{x^{6}}{120} + O(x^{8}) \eqqcolon u \quad(x\to 0)$$
より
\begin{align} e^{x\sin{x}} &= 1+u+\frac{u^{2}}{2} + \frac{u^{3}}{6} +O(u^{4}) \\ &= 1+x^{2}+\left(-\frac{1}{6}+\frac{1}{2}\right)x^{4} + \left(\frac{1}{120}+\frac{1}{2}\left(-\frac{1}{3}\right)+\frac{1}{6}\right)x^{6} +O(x^{8}) \\ &= 1+x^{2}+\frac{1}{3}x^{4}+\frac{1}{120}x^{6}+O(x^{8}) \quad(x\to 0) \end{align}
を得る．

(iv)

$$ \frac{\sin{x}}{x} = 1-\frac{x^{2}}{6}+\frac{x^{4}}{120}-\frac{x^{6}}{7!} +O(x^{8}) \eqqcolon 1-u \quad(x\to 0)$$
より
\begin{align} \frac{x}{\sin{x}} = (1-u)^{-1} &= 1+u+u^{2}+u^{3}+O(u^{4}) \\ &= 1+\frac{1}{6}x^{2}+\left(-\frac{1}{120}+\frac{1}{36}\right)x^{4}+\left(\frac{1}{7!}-\frac{1}{360}+\frac{1}{6^{3}}\right)x^{6} +O(x^{8})\\ &= 1+\frac{1}{6}x^{2} + \frac{7}{320}x^{4}+\left(\frac{1}{7!}-\frac{1}{360}+\frac{1}{6^{3}}\right)x^{6} +O(x^{8}) \quad(x\to 0) \\ &\eqqcolon 1+v \end{align}
となるので，
\begin{align} \log\frac{x}{\sin{x}} = \log(1+v) &= v-\frac{v^{2}}{2}+\frac{v^{3}}{3} +O(v^{4}) \\ &= \frac{1}{6}x^{2} +\left(\frac{7}{320}-\frac{1}{2}\frac{1}{36}\right)x^{4} + \left(\left(\frac{1}{7!}-\frac{1}{360}+\frac{1}{6^{3}}\right)-\frac{1}{2}\frac{7}{3\cdot 360} + \frac{1}{3}\frac{1}{6^{3}}\right)x^{6} +O(x^{8}) \\ &= \frac{1}{6}x^{2} + \frac{1}{180}x^{4} + \frac{1}{2835}x^{6} +O(x^{8}) \quad(x\to 0) \end{align}
が成り立つ．

4

さて，積分型の剰余項
$$ R_{n}(x) = \int_{x_{0}}^{x} \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t) \,\d{t} = \frac{1}{(n-1)!} \int_{x_{0}}^{x} f^{(n)}(t)(x-t)^{n-1} \,\d{t}$$
において，

1. 被積分函数を
   $$ \varphi(t) \coloneqq f^{(n)}(t),\ \psi(t) \coloneqq (x-t)^{n-1}$$
   と分割すると，mvtより，$\xi\in\,]x_{0},x[$であって
   $$ R_{n}(x)=\frac{1}{(n-1)!}\cdot \varphi(\xi) \int_{x_{0}}^{x}\psi(t)\,\d{t} = \frac{f^{(n)}(\xi)}{(n-1)!}\frac{(x-x_{0})^{n}}{n} = \frac{f^{(n)}(\xi)}{n!}(x-x_{0})^{n}$$
   を満たすものが存在する．
2. 被積分函数を
   $$ \varphi(t) \coloneqq f^{(n)}(t)(x-t)^{n-p},\ \psi(t) \coloneqq (x-t)^{p-1},\ p\in [1,n]$$
   と分割すると，mvtより，$\theta\in\,]0,1[$であって
   $$ R_{n}(x) = \frac{1}{(n-1)!} \cdot \varphi(x_{0}+\theta(x-x_{0})) \int_{x_{0}}^{x} \psi(t)\,\d{t} = \frac{f^{(n)}(x_{0}+\theta(x-x_{0}))((1-\theta)(x-x_{0}))^{n-p}}{(n-1)!}\frac{(x-x_{0})^{p}}{p} = \frac{(1-\theta)^{n-p}f^{(n)}(x_{0}+\theta(x-x_{0}))}{p(n-1)!}(x-x_{0})^{n}$$
   を満たすものが存在する．

5

（cf. d-ic, Prob.III.10.）

$$ \frac{1}{k+1} < x \leq \frac{1}{k} \iff k \leq \frac{1}{x} < k+1 \iff k = \left\lfloor \frac{1}{x} \right\rfloor$$
であるから，
\begin{align} a_{n+1} \coloneqq \int_{\frac{1}{n+1}}^{1} \left(\frac{1}{x}-\left\lfloor \frac{1}{x} \right\rfloor\right)\,\d{x} &= \sum_{k=1}^{n} \int_{\frac{1}{k+1}}^{\frac{1}{k}} \left(\frac{1}{x}-\left\lfloor \frac{1}{x} \right\rfloor\right)\,\d{x} \\ &= \sum_{k=1}^{n} \left(\int_{\frac{1}{k+1}}^{\frac{1}{k}} \frac{1}{x}\,\d{x} - \int_{\frac{1}{k+1}}^{\frac{1}{k}} k\,\d{x}\right) \\ &= \sum_{k=1}^{n} \left(\log(k+1)-\log{k}-k\left(\frac{1}{k}-\frac{1}{k+1}\right)\right) \\ &= \sum_{k=1}^{n} \left(\log(k+1)-\log{k}-\frac{1}{k+1}\right) \\ &= \log(n+1)-\sum_{k=1}^{n} \frac{1}{k+1} \\ &= 1 - \left(\sum_{k=1}^{n+1} \frac{1}{k} -\log(n+1)\right)\\ &\to 1 - \gamma \quad(n\to\infty) \end{align}
となる．よって，任意の$\varepsilon\in\,]0,1]$に対して$n \coloneqq \lfloor \varepsilon^{-1} \rfloor \in\mathbb{N}$とおけば
$$ \frac{1}{n+1} < \varepsilon \leq \frac{1}{n} \quad\leadsto\quad a_{n} \leq \int_{\varepsilon}^{1} \left(\frac{1}{x}-\left\lfloor \frac{1}{x} \right\rfloor\right)\,\d{x} \leq a_{n+1}$$
が成り立つことと合わせて，
$$ \int_{0}^{1} \left(\frac{1}{x}-\left\lfloor \frac{1}{x} \right\rfloor\right)\,\d{x} = \lim_{\varepsilon\downarrow 0} \int_{\varepsilon}^{1} \left(\frac{1}{x}-\left\lfloor \frac{1}{x} \right\rfloor\right)\,\d{x} = \lim_{n\to\infty} a_{n} = 1-\gamma$$
を得る．

6

(i)

\begin{align} \int_{k}^{k+1} \log{x}\,\d{x} &= [x\log{x}-x]_{k}^{k+1} \\ &= (k+1)\left(\log{k}+\log \left(1+\frac{1}{k}\right)\right)-k\log{k}-1 \\ &= \log{k} + (k+1)\log \left(1+\frac{1}{k}\right)-1 \\ &= \log{k} + (k+1)\left(\frac{1}{k}+o\left(\frac{1}{k}\right)\right)-1 \\ &= \log{k} + \frac{1}{k} + o(1) \\ &\sim \log{k} \quad(k\to\infty) \end{align}
より
$$ \sum_{k=1}^{n} \log{k} \sim \int_{1}^{n+1} \log{x}\,\d{x} = (n+1)\log(n+1)-n \sim n\log{n} \quad(n\to\infty)$$
となる．

(ii)

$$ \int_{k}^{k+1} (\log{x})^{p}\,\d{x} = (k+1)(\log(k+1))^{p}-k(\log{k})^{p}-p\int_{k}^{k+1} (\log{x})^{p-1}\,\d{x}$$
および
\begin{align} (k+1)(\log(k+1))^{p} &= (k+1)(\log{k})^{p} \left(1+\frac{1}{\log{k}}\log\left(1+\frac{1}{k}\right)\right)^{p} \\ &= (k+1)(\log{k})^{p} \left(1+ \underbrace{\frac{1}{k\log{k}} + o\left(\frac{1}{k\log{k}}\right)}_{\eqqcolon u}\right)^{p} \\ &= (k+1)(\log{k})^{p}(1+pu+ o(u)) \\ &= k(\log{k})^{p}+(\log{k})^{p} + p(\log{k})^{p-1}\left(1+\frac{1}{k}\right) + o((\log{k})^{p-1}) \quad(k\to \infty) \end{align}
より
$$ \int_{k}^{k+1} (\log{x})^{p}\,\d{x} \sim (\log{k})^{p} \quad(k\to\infty)$$
となるので，
$$ \sum_{k=1}^{n} (\log{k})^{p} \sim \int_{1}^{n+1} (\log{x})^{p}\,\d{x} = (n+1)(\log(n+1))^{p} - p\int_{1}^{n+1} (\log{x})^{p-1}\,\d{x} \sim n(\log{n})^{p} \quad(n\to\infty)$$
を得る．

(iii)

$$ \int_{k}^{k+1} \frac{(\log{x})^{p}}{x}\,\d{x} = \frac{1}{p+1}((\log(k+1))^{p+1}-(\log{k})^{p+1}) = \frac{1}{p+1}((\log{k})^{p+1}(1+pu+o(u)) - (\log{k})^{p+1}) = \frac{(\log{k})^{p}}{k}+o\left(\frac{(\log{k})^{p}}{k}\right) \sim \frac{(\log{k})^{p}}{k} \quad(k\to\infty)$$
より
$$ \sum_{k=1}^{n} \frac{(\log{k})^{p}}{k} \sim \int_{1}^{n+1} \frac{(\log{x})^{p}}{x} \,\d{x} = \frac{(\log(n+1))^{p+1}}{p+1} \sim \frac{(\log{n})^{p+1}}{p+1} \quad(n\to\infty)$$
を得る．

7

2(i)と同様にして$f(x)$が$x\to+\infty$のとき無限小であることがわかる．

1. $\exp(-t^{2}/2) = (\exp(-t^{2}/2))'/(-t)$と見て部分積分すると
   $$ \int_{x}^{+\infty} e^{-\frac{t^{2}}{2}}\,\d{t} = \frac{e^{-\frac{x^{2}}{2}}}{x} - \int_{x}^{+\infty} \frac{e^{-\frac{t^{2}}{2}}}{t^{2}}\,\d{t}$$
   となる．また，
   $$ \frac{e^{-\frac{t^{2}}{2}}}{t^{2}} = O\left(\frac{e^{-\frac{t^{2}}{2}}}{t}\right) \quad\leadsto\quad \int_{x}^{+\infty} \frac{e^{-\frac{t^{2}}{2}}}{t^{2}}\,\d{t} = O\left(\int_{x}^{+\infty} \frac{e^{-\frac{t^{2}}{2}}}{t}\,\d{t}\right);\ \int_{x}^{+\infty} \frac{e^{-\frac{t^{2}}{2}}}{t}\,\d{t} = \frac{e^{-\frac{x^{2}}{2}}}{x^{2}} - 2\int_{x}^{+\infty} \frac{e^{-\frac{t^{2}}{2}}}{t^{3}}\,\d{t}$$
   であるから，
   $$ \int_{x}^{+\infty} e^{-\frac{t^{2}}{2}}\,\d{t} = \frac{1}{x}e^{-\frac{x^{2}}{2}} + O\left(\frac{e^{-\frac{x^{2}}{2}}}{x^{2}}\right) \quad(x\to+\infty)$$
   が成り立つ．
2. 部分積分を繰返して
   \begin{align} \int_{x}^{+\infty} e^{-\frac{t^{2}}{2}}\,\d{t} &= \frac{e^{-\frac{x^{2}}{2}}}{x} - \int_{x}^{+\infty} \frac{e^{-\frac{t^{2}}{2}}}{t^{2}}\,\d{t} &&= \frac{1}{x}e^{-\frac{x^{2}}{2}} + O\left(\frac{e^{-\frac{x^{2}}{2}}}{x^{2}}\right) \\ &= \frac{e^{-\frac{x^{2}}{2}}}{x} - \frac{e^{-\frac{x^{2}}{2}}}{x^{3}} + 3\int_{x}^{+\infty} \frac{e^{-\frac{t^{2}}{2}}}{t^{4}}\,\d{t} &&= \left(\frac{1}{x} - \frac{1}{x^{3}}\right)e^{-\frac{x^{2}}{2}} + O\left(\frac{e^{-\frac{x^{2}}{2}}}{x^{4}}\right)\\ &= \frac{e^{-\frac{x^{2}}{2}}}{x} - \frac{e^{-\frac{x^{2}}{2}}}{x^{3}} + 3\frac{e^{-\frac{x^{2}}{2}}}{x^{5}} -5\cdot 3 \int_{x}^{+\infty} \frac{e^{-\frac{t^{2}}{2}}}{t^{6}}\,\d{t} &&= \left(\frac{1}{x}-\frac{1}{x^{3}}+\frac{3}{x^{5}}\right)e^{-\frac{x^{2}}{2}} + O\left(\frac{e^{-\frac{x^{2}}{2}}}{x^{6}}\right) \\ &= \cdots \end{align}
   を得る．

8

(i)

$f(x)\coloneqq x^{2}-7$とおくと，$f'(x)=2x,\,f''(x)=2$であるから，
$$ x_{0} \coloneqq \frac{5}{2},\ c \coloneqq \frac{1}{2},\ \lambda \coloneqq \frac{1}{4},\ \mu \coloneqq 2 \quad\leadsto\quad q = \frac{1}{4}$$
とおけばnewtonが適用できる．このとき
$$ x_{n+1} = \frac{1}{2}\left(x_{n}+\frac{7}{x_{n}}\right),\ |x_{\infty}-x_{n}| \leq 4\left(\frac{1}{8}\right)^{2^{n}}$$
であり（cf. 第1章演習問題3(iii)），
$$ 4\left(\frac{1}{8}\right)^{2^{3}} < \frac{1}{10^{5}} < 4\left(\frac{1}{8}\right)^{2^{2}}$$
であるから，$x_{3}$まで求めればよい：
\begin{align} x_{1} &= \frac{1}{2}\left(\frac{5}{2}+\frac{14}{5}\right) = \frac{53}{20} = 2.65;\\ x_{2} &= \frac{1}{2}\left(\frac{53}{20}+\frac{140}{53}\right) = \frac{5609}{2120} = 2.645754716\cdots;\\ x_{3} &= \frac{1}{2}\left(\frac{5609}{2120}+\frac{14840}{5609}\right) = \textcolor{red}{2.64575}1311\cdots. \end{align}

(ii)

$f(x)\coloneqq x^{3}-25$とおくと，$f'(x)=3x^{2},\,f''(x)=6x$であるから，
$$ x_{0} \coloneqq 3,\ c \coloneqq \frac{1}{6},\ \lambda \coloneqq \frac{1}{24},\ \mu \coloneqq 17 \quad\leadsto\quad q = \frac{17}{48}$$
とおけばnewtonが適用できる．このとき
$$ x_{n+1} = \frac{1}{3}\left(2x_{n}+\frac{25}{x_{n}^{2}}\right),\ |x_{\infty}-x_{n}| \leq \frac{48}{17}\left(\frac{17}{288}\right)^{2^{n}}$$
であり
$$ \frac{48}{17}\left(\frac{17}{288}\right)^{2^{3}} < \frac{1}{10^{5}} < \frac{48}{17}\left(\frac{17}{288}\right)^{2^{2}}$$
であるから，$x_{3}$まで求めればよい：
\begin{align} x_{1} &= \frac{1}{3}\left(6+\frac{25}{9}\right) = \frac{79}{27} = 2.925925\cdots;\\ x_{2} &= \frac{1}{3}\left(\frac{158}{27}+\frac{18225}{6241}\right) = \frac{1478153}{505521} = 2.924018982\cdots;\\ x_{3} &= \frac{1}{3}\left(\frac{2956306}{505521}+\frac{25\times 505521^{2}}{1478153^{2}}\right) = \textcolor{red}{2.92401}7738\cdots. \end{align}

(iii)

\begin{align} x_{1} &= \frac{1}{2}\left(9+\frac{79}{9}\right) = \frac{80}{9} = 8.88888888\cdots;\\ x_{2} &= \frac{1}{2}\left(\frac{80}{9}+\frac{711}{80}\right) = \frac{12799}{1440} = 8.888194444\cdots;\\ x_{3} &= \frac{1}{2}\left(\frac{12799}{1440}+\frac{79 \times 1440}{12799}\right) = \textcolor{red}{8.88819}4417\cdots. \end{align}

(iv)

\begin{align} \sqrt[3]{2} &= \left(\left(\frac{5}{4}\right)^{3}+2-\left(\frac{5}{4}\right)^{3}\right)^{\frac{1}{3}} \\ &= \frac{5}{4}\left(1+\frac{4^{3}}{5^{3}}\left(2-\frac{5^{3}}{4^{3}}\right)\right)^{\frac{1}{3}} \\ &= \frac{5}{4}\left(1+\frac{3}{125}\right)^{\frac{1}{3}} \\ &= \frac{5}{4}\left(1+\frac{1}{125}-\frac{1}{9}\left(\frac{3}{125}\right)^{2} + \frac{5}{81}\left(\frac{3}{125}\right)^{3} +\cdots \right) \\ &= 1.25 + \frac{5}{4}\left(\frac{1}{125}-\frac{1}{9}\left(\frac{3}{125}\right)^{2} + \frac{5}{81}\left(\frac{3}{125}\right)^{3} +\cdots \right) \\ &= 1.26 + \frac{5}{4}\left(-\frac{1}{9}\left(\frac{3}{125}\right)^{2} + \frac{5}{81}\left(\frac{3}{125}\right)^{3} +\cdots \right) \\ &= 1.25992 + \frac{5}{4}\left(\frac{5}{81}\left(\frac{3}{125}\right)^{3}+\cdots \right) \\ &= \textcolor{red}{1.25992}5333\cdots + \frac{5}{4}\left(\binom{\frac{1}{3}}{4}\left(\frac{3}{125}\right)^{4} + \cdots \right). \end{align}

9

\begin{align} \log{2} &= 2\log\frac{3}{2}-\log\frac{9}{8};\\ \log{3} &= 3\log\frac{3}{2}-\log\frac{9}{8}; \end{align}
という仕組．なのだが
$$ \log\frac{x+1}{x} = \log \left(1+\frac{1}{x}\right) = \frac{1}{x}-\frac{1}{2x^{2}}+\frac{1}{3x^{3}}-\frac{1}{4x^{4}}+\cdots $$
は収束が遅くて困っていたところ，以下を見つけた：

rapid-convより
\begin{align} \log\frac{3}{2} &= 2\left(\frac{1}{5} + \frac{1}{3\cdot 5^{3}} + \frac{1}{5\cdot 5^{5}} + \frac{1}{7\cdot 5^{7}} + \frac{1}{9\cdot 5^{9}} +\cdots{} \right) \\ &= 0.4 + 2\left(\frac{1}{3\cdot 5^{3}} + \frac{1}{5\cdot 5^{5}} + \frac{1}{7\cdot 5^{7}} + \frac{1}{9\cdot 5^{9}} +\cdots{} \right) \\ &= 0.405333\cdots + 2\left(\frac{1}{5\cdot 5^{5}} + \frac{1}{7\cdot 5^{7}} + \frac{1}{9\cdot 5^{9}} +\cdots{} \right) \\ &= 0.405461333\cdots + 2\left(\frac{1}{7\cdot 5^{7}} + \frac{1}{9\cdot 5^{9}} +\cdots{} \right) \\ &= \textcolor{red}{0.40546}49904\cdots + 2\left(\frac{1}{9\cdot 5^{9}} +\cdots{} \right) \\ &\\ \log\frac{9}{8} &= 2\left(\frac{1}{17} + \frac{1}{3\cdot 17^{3}} + \frac{1}{5\cdot 17^{5}} + \frac{1}{7\cdot 17^{7}}+\cdots{}\right) \\ &= 0.117647\cdots +2\left(\frac{1}{3\cdot 17^{3}} + \frac{1}{5\cdot 17^{5}} + \frac{1}{7\cdot 17^{7}}+\cdots{}\right) \\ &= 0.1177827\cdots +2\left(\frac{1}{5\cdot 17^{5}} + \frac{1}{7\cdot 17^{7}}+\cdots{}\right) \\ &= \textcolor{red}{0.11778}30\cdots +2\left(\frac{1}{7\cdot 17^{7}}+\cdots{}\right) \\ \end{align}
となるので，
$$ \log{2} \approx 0.69314,\ \log{3} \approx 1.0986$$
を得る．

13

$$ \int (\sin{x})e^{-x}\,\d{x} = -\frac{1}{2}(\cos{x}+\sin{x})e^{-x}$$
より
\begin{align} \int_{2n\pi}^{(2n+1)\pi} (\sin{x})e^{-x}\,\d{x} &= \frac{1}{2}(e^{-(2n+1)\pi}+e^{-2n\pi})=\frac{e^{-\pi}+1}{2}e^{-2n\pi};\\ \int_{(2n+1)\pi}^{(2n+2)\pi} (-\sin{x})e^{-x}\,\d{x} &= \frac{1}{2}(e^{-(2n+2)\pi}+e^{-(2n+1)\pi}) = \frac{e^{-\pi}+1}{2}e^{-(2n+1)\pi}; \end{align}
が成り立つ．よって
$$ \int_{0}^{\infty} |\sin{x}|e^{-x}\,\d{x} = \frac{e^{-\pi}+1}{2} \sum_{n=0}^{\infty} e^{-n\pi} = \frac{e^{-\pi}+1}{2} \frac{1}{1-e^{-\pi}} = \frac{1}{2} \frac{e^{\frac{\pi}{2}}+e^{-\frac{\pi}{2}}}{e^{\frac{\pi}{2}}-e^{-\frac{\pi}{2}}} = \frac{1}{2}\coth\frac{\pi}{2}$$
を得る．

14

点列
$$ a_{n} \coloneqq \left(\sum_{k=1}^{n} \frac{(-1)^{k-1}}{k},0\right)$$
は収束するが（cf. p.111, 例2），
$$ \sum_{n=1}^{\infty} |a_{n+1}-a_{n}| = \sum_{n=1}^{\infty} \frac{1}{n+1} = +\infty$$
となる．

15

(i)

$a=b$のときは明らかに発散する．$a\neq b$のとき，
$$ \frac{a_{n}}{a_{n-1}} = \frac{na+1}{nb+1} \to \frac{a}{b}$$
となるので，ダランベールの判定法より$a< b$のとき収束し$a>b$のとき発散する．

(ii)

（cf. ガウスの超幾何関数 ）

$$ \frac{a_{n}}{a_{n-1}} = \frac{(\alpha+n)(\beta+n)}{(n+1)(\gamma+n)} = \frac{1+\frac{\alpha+\beta}{n}+\frac{\alpha\beta}{n^{2}}}{1+\frac{\gamma+1}{n}+\frac{\gamma}{n^{2}}} \to 1 \quad(n\to\infty)$$
なのでダランベールの判定法は使えないが，
$$ \frac{a_{n}}{a_{n-1}} - \left(1-\frac{(\gamma+1)-(\alpha+\beta)}{n}\right) = \frac{\frac{\alpha\beta-\gamma+(\gamma+1)((\gamma+1)-(\alpha+\beta))}{n^{2}} + \frac{\gamma((\gamma+1)-(\alpha+\beta))}{n^{3}}}{1+\frac{\gamma+1}{n}+\frac{\gamma}{n^{2}}}$$
より
$$ \frac{a_{n}}{a_{n-1}} = 1-\frac{\gamma-(\alpha+\beta)+1}{n} + O\left(\frac{1}{n^{2}}\right) \quad(n\to\infty)$$
となるので，ガウスの判定法より，$\gamma>\alpha+\beta$のとき収束し$\gamma\leq \alpha+\beta$のとき発散する．

(iii)

$n\to\infty$のとき
\begin{align} \log(n^{2}-1) &= 2\log{n}+\log \left(1-\frac{1}{n^{2}}\right) = 2\log{n}-\frac{1}{n^{2}}+o\left(\frac{1}{n^{2}}\right);\\ \log(n^{3}+1) &= 3\log{n} + \log \left(1+\frac{1}{n^{3}}\right) = 3\log{n}+\frac{1}{n^{3}} +o\left(\frac{1}{n^{3}}\right); \end{align}
となるので，
$$ \frac{(n^{2}-1)\log(n^{2}-1)}{(n^{3}+1)\log(n^{3}+1)} \sim \frac{2}{3n} \quad(n\to\infty)$$
が得られ，したがって発散する．

(iv)

$$ \frac{n^{2}+(\log\log{n})^{8}}{\sqrt{n^{6}-1}(\log{n})^{2}+\log{n}} \sim \frac{n^{2}(1+o(1))}{n^{3}(\log{n})^{2}(1+o(1))} \sim \frac{1}{n(\log{n})^{2}} \quad(n\to\infty).$$

(v)

$$ \frac{\sqrt{n^{2}+1}\log(n^{2}+1) + \sqrt{n^{2}-1}\log(n^{2}-1)}{n^{2}(\log{n})^{2}(\log\log{n})^{2}} \sim \frac{4n\log{n}}{n^{2}(\log{n})^{2}(\log\log{n})^{2}} = \frac{4}{n\log{n}(\log\log{n})^{2}} \quad(n\to\infty)$$
であり，
$$ \int_{3}^{+\infty} \frac{1}{x\log{x}(\log\log{x})^{2}}\,\d{x} = \int_{\log\log{3}}^{+\infty} \frac{1}{t^{2}}\,\d{t} = \frac{1}{\log\log{3}} <+\infty$$
であるから，収束する（命題3.22）．

(vi)

$$ \frac{\prod_{k=1}^{n}(k^{2}+1)}{n!\,(n+1)!\,(\log{n})^{p}} = \frac{\prod_{k=1}^{n}(k^{2}+1)}{(n!)^{2} \cdot (n+1)(\log{n})^{p}} = \prod_{k=1}^{n} \left(1+\frac{1}{k^{2}}\right) \frac{1}{(n+1)(\log{n})^{p}} \sim \frac{1}{n(\log{n})^{p}} \quad(n\to\infty).$$

16

（cf. d-ic, Prob.III.13.）

$$ a_{n} = (1+a+b)\log{n}+\frac{a+2b}{n}-\frac{a+4b}{2n^{2}}+o\left(\frac{1}{n^{2}}\right) \quad(n\to\infty).$$

17

$p>1$のとき
$$ \sum_{k=1}^{n}\frac{1}{k} \cdot \frac{\sin{n\theta}}{(\log{n})^{p}} \sim \frac{\sin{n\theta}}{(\log{n})^{p-1}} \quad(n\to\infty)$$
であり，
$$ \frac{1}{(\log{n})^{p-1}} > \frac{1}{(\log(n+1))^{p-1}} \to 0 \quad(n\to\infty)$$
であるから，p.115, 例4より$\sum (\sin{n\theta})/(\log{n})^{p-1}$は収束する．

18

(i)

（cf. オストロウスキの定理 ）

$\varepsilon>0$とする．このとき，$n_{0}\in\mathbb{N}$であって
$$ \forall n>n_{0},\ a_{n_{0}+1} +\cdots+ a_{n} < \varepsilon \quad\leadsto\quad (n-n_{0})a_{n} < \varepsilon$$
が成り立つものが存在する．また，$a_{n}\to 0$より，$n_{1}\in\mathbb{N}$であって
$$ \forall n > n_{1},\ a_{n} < \frac{\varepsilon}{n_{0}}$$
が成り立つものが存在する．よって
$$ \forall n >n_{0}+n_{1},\ na_{n} < \varepsilon+n_{0}a_{n} < 2\varepsilon$$
が成り立つ．

(ii)

$2s>1$より$\sum n^{-2s}$は収束することに注意する．このとき，コーシーの不等式（cf. 第2章演習問題8）より
$$ \sum_{n=1}^{N} \frac{|a_{n}|}{n^{s}} \leq \left(\sum_{n=1}^{N} a_{n}^{2}\right)^{\frac{1}{2}} \cdot \left(\sum_{n=1}^{N} \frac{1}{n^{2s}}\right)^{\frac{1}{2}} \leq \left(\sum_{n=1}^{\infty} a_{n}^{2}\right)^{\frac{1}{2}} \cdot \left(\sum_{n=1}^{\infty} \frac{1}{n^{2s}}\right)^{\frac{1}{2}} <+\infty$$
が成り立つので，結論を得る．

(iii)

$n\in\mathbb{N}$が十分大きいとき
$$ a_{n} = \sum_{k=1}^{n}(a_{k}-a_{k-1})+a_{0} \geq n\delta +a_{0} > 0$$
となるので，
$$ \frac{1}{a_{n}^{2}} \leq \frac{1}{(n\delta+a_{0})^{2}} \sim \frac{1}{\delta^{2}}\frac{1}{n^{2}} \quad(n\to\infty)$$
より結論を得る（cf. 定理3.19）．

(iv)

$$ \sum a_{n} = \sum (na_{n})\frac{1}{n}.$$

19

$$ a_{n}b_{n} = a_{n}(b_{n}-b_{1}) + a_{n}b_{1},\ b_{n}-b_{1}\geq 0,\ \sum a_{n}b_{1}:\text{convergent}$$
であるから，$b_{n}\geq 0$としてよい．

$\varepsilon>0$とする．仮定より，$n_{0}\in\mathbb{N}$であって
$$ n \geq n_{0},p \geq 1 \implies \left|\sum_{k=n+1}^{n+p} a_{k}\right| < \varepsilon$$
を満たすものが存在する．このとき（定理3.32の証明中の記号を用いると），$n \geq n_{0},p \geq 1$ならば
\begin{align} |a_{n+1}b_{n+1} +\cdots+ a_{n+p}b_{n+p}| &= |\sigma_{1}(b_{n+1}-b_{n+2}) + \cdots+ \sigma_{p-1}(b_{n+p-1}-b_{n+p}) + \sigma_{p}b_{n+p}| \\ &\leq |\sigma_{1}|(b_{n+2}-b_{n+1}) +\cdots+ |\sigma_{p-1}|(b_{n+p}-b_{n+p-1}) + |\sigma_{p}|b_{n+p} \\ &\leq \varepsilon(b_{n+p}-b_{n+1})+\varepsilon b_{n+p} \\ &\leq 2K\varepsilon \end{align}
が成り立つ．

20

$\ell>1$のとき，$p \in\,]1,\ell[$を取ると，$n_{0}\in\mathbb{N}$であって
$$ n > n_{0} \implies p < n\log\frac{a_{n}}{a_{n+1}} \quad\leadsto\quad \frac{a_{n+1}}{a_{n}} < e^{-\frac{p}{n}}$$
を満たすものが存在する．よって
$$ a_{n+1} < a_{n}e^{-\frac{p}{n}} < a_{n-1}e^{-\frac{p}{n-1}}e^{-\frac{p}{n}} <\cdots< a_{1}\exp \left(-p\sum_{k=1}^{n}\frac{1}{k}\right) \sim \frac{a_{1}}{e^{p\gamma}}\frac{1}{n^{p}} \quad(n\to\infty)$$
が成り立つ（ただし$\gamma$はオイラーの定数である）．

23

いま，$Q_{1},Q_{2}$は絶対収束なので
$$ Q_{1}Q_{2} = \prod (1+a^{2n})(1+a^{2n-1}) = \prod(1+a^{m})$$
が成り立つ．また，
$$ (1-a^{2k-1})\underbrace{(1+a^{2k-1})(1+a^{(2k-1)2})\cdots(1+a^{(2k-1)2^{N}})}_{\eqqcolon p_{k,N}} = 1-a^{(2k-1)2^{N+1}} \to 1 \quad(N\to\infty)$$
より，$b_{k}\coloneqq \lim p_{k,N}$とおき$Q_{3}$の部分積を$q_{n}$とおくと，
$$ Q_{1}Q_{2}q_{n} = Q_{1}Q_{2}(1-a)(1-a^{3})\cdots(1-a^{2n-1}) = \frac{\prod(1+a^{m})}{b_{1}b_{2}\cdots b_{n}}$$
となる．ところで，$n\in\mathbb{N}$に対して，十分大きな$N,m$を取ると
$$ 1 \leq \frac{(1+a)\cdots(1+a^{m})}{p_{1,N}\cdots p_{n,N}} \leq (1+a^{2n+1})\cdots(1+a^{m})$$
が成り立つことから，
$$ 1 \leq \frac{\prod(1+a^{m})}{p_{1,N}\cdots p_{n,N}} \leq \frac{\prod(1+a^{m})}{(1+a)\cdots(1+a^{2n})} \quad\leadsto\quad 1 \leq \frac{\prod(1+a^{m})}{b_{1}\cdots b_{n}} \leq \frac{\prod(1+a^{m})}{(1+a)\cdots(1+a^{2n})}$$
となるので，$n\to\infty$として
$$ 1 \leq Q_{1}Q_{2}Q_{3} \leq 1$$
を得る．

24

（cf. Euler, Various observations on angles proceeding in geometric progression ）

$0< a_{1} = 1/\sqrt{2}<1$であり
$$ 0< a_{n}<1 \implies 0 < \sqrt{\frac{1}{2}} < \sqrt{\frac{1+a_{n}}{2}} = a_{n+1} < 1$$
が成り立つので，$\theta_{n} \in \,]0,\pi/2[$であって$a_{n}=\cos\theta_{n}$なるものがただ一つ存在する．さらに，$\theta\coloneqq \pi/2$とおくと
$$ a_{0} = 0 = \cos\theta$$
であり，
$$ \cos\theta_{n+1} = \sqrt{\frac{1+\cos\theta_{n}}{2}} \quad\leadsto\quad \cos\theta_{n} = 2(\cos\theta_{n+1})^{2}-1 = \cos(2\theta_{n+1}),\ \theta_{n},2\theta_{n+1} \in [0,\pi]$$
より
$$ \theta_{n}=\frac{\theta}{2^{n}} \implies \theta_{n+1}=\frac{\theta}{2^{n+1}}$$
が成り立つので，結局
$$ p_{n} = a_{1}\cdots a_{n} = \cos\frac{\theta}{2} \cdots \cos\frac{\theta}{2^{n}}$$
となる．ところで，
$$ p_{n}\sin\frac{\theta}{2^{n}} = \frac{p_{n-1}}{2}\sin\frac{\theta}{2^{n-1}} = \frac{p_{n-2}}{2^{2}}\sin\frac{\theta}{2^{n-2}} =\cdots= \frac{p_{1}}{2^{n-1}}\sin\frac{\theta}{2} = \frac{\sin\theta}{2^{n}}$$
が成り立つので，
$$ p_{n} = \frac{\sin\theta}{2^{n}\sin\frac{\theta}{2^{n}}} = \frac{\sin\theta}{\theta} \frac{\frac{\theta}{2^{n}}}{\sin\frac{\theta}{2^{n}}} \to \frac{\sin\theta}{\theta} = \frac{2}{\pi} \quad(n\to\infty)$$
を得る．