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$n\in\N_+$
$\beginend{align}{
\hat\lambda(n) \acoloneqq
\sum_{k=0}^\infty \lr({\frac{(-1)^k}{2k+1}})^n \\&=
\frac{A_{n-1}}{2(n-1)!}\lr({\frac\pi2})^n
}$
$\beginend{align}{
\sum_{n=1}^\infty \hat\lambda(n)z^n &=
\sum_{n=1}^\infty
z^n \sum_{k=0}^\infty \lr({\frac{(-1)^k}{2k+1}})^n =
\sum_{k=0}^\infty
\frac{\frac{(-1)^kz}{2k+1}}{1-\frac{(-1)^kz}{2k+1}} \\&=
z\sum_{k=0}^\infty \frac{(-1)^k}{2k+1+(-1)^{k+1}z} =
z\sum_{k=0}^\infty \lr({\frac1{4k+1-z}-\frac1{4k+3+z}}) \\&=
z\sum_{k=0}^\infty \frac{2(z+1)}{(4k+2)^2-(z+1)^2} =
\frac z4\sum_{k=0}^\infty
\frac{\frac{z+1}2}{\lr({k+\frac12})^2-\lr({\frac{z+1}4})^2} \\&\stackrel※=
\frac{\pi z}4\tan{\frac{\pi(z+1)}4} =
\frac{\pi z}4\sum_{n=0}^\infty \frac{A_n}{n!}\lr({\frac{\pi z}2})^n =
\frac12\sum_{n=0}^\infty \frac{A_n}{n!}\lr({\frac{\pi z}2})^{n+1} \\
\sahen &= \uhen
}$
$※\because$
三角関数の部分分数展開$\tan$
$\hat\lambda(n) = \beginend{cases}{
\lr({1-2^{-n}})\zeta(n) & n \equiv 0 \\
\beta{(n)} & n \equiv 1
} \pmod 2$であるため、
リーマンゼータ関数の偶数値とディリクレベータ関数の奇数値が求まりました。
$A_n$の値、
A000111 - OEIS
より。
| $n$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
|---|
| $A_n$ | $1$ | $1$ | $1$ | $2$ | $5$ | $16$ | $61$ | $272$ |
