交代級数にさらに極限をつける（階乗の逆数交代和）

$f_n(t)={\displaystyle \frac{(2n+1{)}^t(2n+2{)}^t-(2n+2{)}^t}{(2n+1{)}^t(2n+2{)}^t-1}}$ と変形できる.

よって任意の正実数$t$について,

$g(x):={\displaystyle \frac{x^t(x+1{)}^t-(x+1{)}^t}{x^t(x+1{)}^t-1}}$ が $x\ge 1$ で単調増加であることを示せば良い.

$g^{\prime }(x)={\displaystyle \frac{t(x+1{)}^t\{((x+1{)}^t+x^{-t}-2)x^{t+1}+((x+1{)}^t-1)x^t\}}{x(x^t(x+1{)}^t-1{)}^2}}$

$g^{\prime }(x)$の分子に着目すると,
$(x+1{)}^t+x^{-t}-2>0$, $(x+1{)}^t-1>0$ なので,
$g^{\prime }(x)>0$

以上, 証明終わり.

上で述べたように, 左辺は収束する.

${\displaystyle \sum _{n=0}^{\infty }(-1{)}^n{a_n}^t-f_N(t)}$

$={\displaystyle \sum _{n=0}^{\infty }\{f_n(t)({a_{2n}}^t-{a_{2n+2}}^t)\}-f_N(t)}$

$={\displaystyle \sum _{n=0}^{\infty }\{(f_n(t)-f_N(t))({a_{2n}}^t-{a_{2n+2}}^t)\}}$

$>{\displaystyle \sum _{n=N+1}^{\infty }\{(f_{N+1}(t)-f_N(t))({a_{2n}}^t-{a_{2n+2}}^t)\}-{\displaystyle \sum _{n=0}^N\{f_N(t)({a_{2n}}^t-{a_{2n+2}}^t)\}}}$

$=(f_{N+1}(t)-f_N(t)){a_{2N+2}}^t-f_N(t)(1-{a_{2N+2}}^t)$

ただし, 最後の不等式には補題2をもちいた.
最後の式は$t\to +0$で
$\underset{t\to +0}{lim}(f_{N+1}(t)-f_N(t))$
$={\displaystyle \frac{1}{1+\frac{\ln (2N+4)}{\ln (2N+3)}}}-{\displaystyle \frac{1}{1+\frac{\ln (2N+2)}{\ln (2N+1)}}}>0$

に収束するので,
以上が証明となる.

$f_n(t)<{\displaystyle \frac{1-\frac{1}{(2n+1{)}^t}}{1-\frac{1}{(2n+1{)}^t}\frac{1}{(2n+1{)}^t}}}={\displaystyle \frac{1}{1+\frac{1}{(2n+1{)}^t}}}$ である.

${\displaystyle \frac{1}{1+\frac{1}{(2n+1{)}^t}}}\le p(t)$ を$n$について解くと

$2n+1\le q(t)$ となる.

この不等式を満たす整数$n$のうち, 最大のものは$r(t)$である.

よって、$n\le r(t)⟹f_n(t)<p(t)$ が成り立つ.

これにより,

${\displaystyle \sum _{n=0}^{\infty }(-1{)}^n{a_n}^t}$

$={\displaystyle \sum _{n=0}^{\infty }\{f_n(t)({a_{2n}}^t-{a_{2n+2}}^t)\}}$

$={\displaystyle \sum _{n=0}^{r(t)}\{f_n(t)({a_{2n}}^t-{a_{2n+2}}^t)\}+{\displaystyle \sum _{n=r(t)+1}^{\infty }\{f_n(t)({a_{2n}}^t-{a_{2n+2}}^t)\}}}$

$<{\displaystyle \sum _{n=0}^{r(t)}\{p(t)({a_{2n}}^t-{a_{2n+2}}^t)\}+{\displaystyle \sum _{n=r(t)+1}^{\infty }({a_{2n}}^t-{a_{2n+2}}^t)}}$

$=p(t)(1-{a_{2r(t)+2}}^t)+{a_{2r(t)+2}}^t$

以上, 証明終わり.

${a_{2r(t)+2}}^t$

$={\left\{{\displaystyle \frac{1}{(2r(t)+2)!}}\right\}}^t$

$\le {\displaystyle \frac{1}{e^t}}{\left({\displaystyle \frac{e}{2r(t)+2}}\right)}^{(2r(t)+2)t}(\because e{\left({\displaystyle \frac{n}{e}}\right)}^n\le n!)$

$={\displaystyle \frac{1}{e^t}}{\left({\displaystyle \frac{e}{2r(t)+2}}\right)}^{(2r(t)+2)t}$

ここで, $\underset{t\to +0}{lim}(2r(t)+2)t$ を先に求めておく.

$(2r(t)+2)t=(2\left[{\displaystyle \frac{q(t)-1}{2}}\right]+2)t>(q(t)-1)t$ より,

$\underset{t\to +0}{lim}(2r(t)+2)t$

$\ge \underset{t\to +0}{lim}(q(t)-1)t$

$=\underset{t\to +0}{lim}\{\exp (-{\displaystyle \frac{1}{t\ln (t)}})-1\}t$

$=\underset{u\to +\mathrm{\infty }}{lim}\{{\displaystyle \frac{1}{u}}\exp \left({\displaystyle \frac{u}{\ln (u)}}\right)-{\displaystyle \frac{1}{u}}\}(u={\displaystyle \frac{1}{t}}\text{と変換した.})$

$=+\mathrm{\infty }$

以上より,

$\underset{t\to +0}{lim}{a_{2r(t)+2}}^t$

$\le \underset{t\to +0}{lim}{\displaystyle \frac{1}{e^t}}{\left({\displaystyle \frac{e}{2r(t)+2}}\right)}^{(2r(t)+2)t}$

$=1\cdot 0^{\mathrm{\infty }}=0$

以上, 証明終わり.

補題4と補題5より,任意の自然数$N$について

$\underset{t\to +0}{lim}{\displaystyle \frac{{a_{2N}}^t-{a_{2N+1}}^t}{{a_{2N}}^t-{a_{2N+2}}^t}}\le \underset{t\to +0}{lim}\left\{{\displaystyle \sum _{n=0}^{\infty }{\displaystyle \frac{(-1{)}^n}{(n!{)}^t}}}\right\}\le \underset{t\to +0}{lim}\{p(t)(1-{a_{2r(t)+2}}^t)+{a_{2r(t)+2}}^t\}$

左辺,右辺の極限を調べる.
まずは左辺から

$\underset{t\to +0}{lim}{\displaystyle \frac{{a_{2N}}^t-{a_{2N+1}}^t}{{a_{2N}}^t-{a_{2N+2}}^t}}$

$={\displaystyle \frac{\ln \left(a_{2N}\right)-\ln \left(a_{2N+1}\right)}{\ln \left(a_{2N}\right)-\ln \left(a_{2N+2}\right)}}$

$={\displaystyle \frac{\ln (2N+1)}{\ln \{(2N+1)(2N+2)\}}}$

これが任意の$N$で成り立つことと,

$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{\ln (2N+1)}{\ln \{(2N+1)(2N+2)\}}}={\displaystyle \frac{1}{2}}$ であることから,

${\displaystyle \frac{1}{2}}\le \underset{t\to +0}{lim}\left\{{\displaystyle \sum _{n=0}^{\infty }{\displaystyle \frac{(-1{)}^n}{(n!{)}^t}}}\right\}$

次に右辺を調べる

$\underset{t\to +0}{lim}p(t)={\displaystyle \frac{1}{2}}$

$\underset{t\to +0}{lim}{a_{2r(t)+2}}^t=0$

以上より, 右辺も${\displaystyle \frac{1}{2}}$に収束する.

以上, 証明終わり.