多项式倒数 求和

hbghlyj

\begin{aligned}
\sum_{n=0}^{\infty} \frac{1}{n^2+n+k}&=\pi \frac{\tan (\sqrt{1-4 k}\frac \pi 2)}{\sqrt{1-4 k}}\\
\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)(2 n-1)}&=\log (4)-1 \\
\sum_{n=1}^{\infty} \frac{(-1)^n}{n(2 n+1)(2 n-1)}&=\log (2)-1
\end{aligned}

kuing

第一个和以前这帖有点像：forum.php?mod=viewthread&tid=1668
和这个也不知有没有关联：forum.php?mod=viewthread&tid=7822

tommywong

答咗一個先

en.wikipedia.org/wiki/Mittag-Leffler's_theorem

$\displaystyle \tan(z) = \sum_{n=0}^\infty \frac{8z}{(2n+1)^2\pi^2-4z^2}$
$\displaystyle \sum_{n=0}^\infty \frac{1}{(n+\dfrac{1}{2})^2-x^2}=\dfrac{\pi\tan(\pi x)}{2x}$

Put $~x=\dfrac{\sqrt{1-4k}}{2}$
$\displaystyle \sum_{n=0}^{\infty} \frac{1}{n^2+n+k}=\pi \frac{\tan (\sqrt{1-4 k}\frac \pi 2) }{\sqrt{1-4 k}}$

abababa

第二个n=0时没有定义吧，觉得是从n=1开始
\[\sum_{n=1}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n}\right)+\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\ln2-1+\ln2=2\ln2-1\]

hbghlyj

abababa 发表于 2023-8-14 12:42
第二个n=0时没有定义吧，觉得是从n=1开始
\[\sum_{n=1}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n}\right) ...

已修改。
谢谢。原来就是log级数啊。

hbghlyj

Make a Taylor series
$$  \sum_{n=1}^{\infty} \frac{x^n}{n(2n+1)(2n-1)} = \log(1-x) + (1+x) {\rm arctanh}(\sqrt{x})/\sqrt{x} $$
Take limit $x \to 1$.

tommywong

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{n(2n+1)(2n-1)}
=\sum_{n=1}^{\infty} \left(\frac{(-1)^n}{2n+1}+\dfrac{(-1)^n}{2n-1}\right)-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=-1+\ln 2$

hbghlyj

\begin{aligned}
& \csc z=\frac{1}{z}-2 z\left(\frac{1}{z^2-\pi^2}-\frac{1}{z^2-4 \pi^2}+\frac{1}{z^2-9 \pi^2}-\cdots\right) \\
& \sec z=\pi\left(\frac{1}{(\pi / 2)^2-z^2}-\frac{3}{(3 \pi / 2)^2-z^2}+\frac{5}{(5 \pi / 2)^2-z^2}-\cdots\right) \\
& \tan z=2 z\left(\frac{1}{(\pi / 2)^2-z^2}+\frac{1}{(3 \pi / 2)^2-z^2}+\frac{1}{(5 \pi / 2)^2-z^2}+\cdots\right) \\
& \cot z=\frac{1}{z}+2 z\left(\frac{1}{z^2-\pi^2}+\frac{1}{z^2-4 \pi^2}+\frac{1}{z^2-9 \pi^2}+\cdots\right) \\
& \operatorname{csch} z=\frac{1}{z}-2 z\left(\frac{1}{z^2+\pi^2}-\frac{1}{z^2+4 \pi^2}+\frac{1}{z^2+9 \pi^2}-\cdots\right) \\
& \operatorname{sech} z=\pi\left(\frac{1}{(\pi / 2)^2+z^2}-\frac{3}{(3 \pi / 2)^2+z^2}+\frac{5}{(5 \pi / 2)^2+z^2}-\cdots\right) \\
& \tanh z=2 z\left(\frac{1}{z^2+(\pi / 2)^2}+\frac{1}{z^2+(3 \pi / 2)^2}+\frac{1}{z^2+(5 \pi / 2)^2}+\cdots\right) \\
& \operatorname{coth} z=\frac{1}{z}+2 z\left(\frac{1}{z^2+\pi^2}+\frac{1}{z^2+4 \pi^2}+\frac{1}{z^2+9 \pi^2}+\cdots\right)
\end{aligned}