求和$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty\frac{1}{k2^n+1}$

elim

题：求和 $\displaystyle\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty\frac{1}{k2^n+1}.$

王守恩

数学万花筒(您会怀疑自己的眼睛吗)。

$\displaystyle\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{1}{2^n*k^2}=\frac{\pi^2}{3}$
$\displaystyle\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{1}{3^n*k^2}=\frac{\pi^2}{4}$
$\displaystyle\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{1}{6^n*k^2}=\frac{\pi^2}{5}$
$\displaystyle\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{1}{2^n*k^2}=\frac{\pi^2}{6}$
$\displaystyle\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{(-1)^n}{6^n*k^2}=\frac{\pi^2}{7}$
$\displaystyle\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{(-1)^n}{3^n*k^2}=\frac{\pi^2}{8}$
$\displaystyle\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{(-1)^n}{2^n*k^2}=\frac{\pi^2}{9}$

$\displaystyle\cdots\cdots$

$\small\displaystyle\sum_{k=1}^\infty\sum_{n=2}^\infty\frac{(-1)^{n}}{2^n*k^2}= \sum_{k=1}^\infty\sum_{n=2}^\infty\frac{1}{3^n*k^2}= \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{k+1}}{4^n*k^2}= \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+1}}{5^n*k^2}= \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{1}{7^n*k^2}=\frac{\pi^2}{36}$

王守恩

接2楼。

$\small\displaystyle\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{1}{2^n*k^2}=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{(-1)^{k-1}}{2^n*k^2}=\bigg(\sum_{k=1}^\infty\frac{1}{k^2}\bigg)\frac{2^{\infty}-1}{2^{\infty}}=\sum_{k=1}^\infty\frac{1}{k^2}=\frac{\pi^2}{6}$

接主帖。

$\small\displaystyle\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty\frac{1}{2^n*k+1}=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{(-1)^{k-1}}{2^n*k^2+k}=\bigg(\sum_{k=1}^\infty\frac{1}{k^2+k}\bigg)\frac{2^{\infty}-1}{2^{\infty}}=\sum_{k=1}^\infty\frac{1}{k^2+k}=1$

ccmmjj

寻找数据规律，是守恩兄的强项。

王守恩

从简单说起,  解锁去枷很有必要。

$\displaystyle\frac{6}{2*1^2}+\frac{6}{2*2^2}+\frac{6}{2*3^2}+\frac{6}{2*4^2}+\frac{6}{2*5^2}+\frac{6}{2*6^2}+\displaystyle\cdots\cdots=\frac{\pi^2}{2}$
$\displaystyle\frac{6}{3*1^2}+\frac{6}{3*2^2}+\frac{6}{3*3^2}+\frac{6}{3*4^2}+\frac{6}{3*5^2}+\frac{6}{3*6^2}+\displaystyle\cdots\cdots=\frac{\pi^2}{3}$
$\displaystyle\frac{6}{4*1^2}+\frac{6}{4*2^2}+\frac{6}{4*3^2}+\frac{6}{4*4^2}+\frac{6}{4*5^2}+\frac{6}{4*6^2}+\displaystyle\cdots\cdots=\frac{\pi^2}{4}$
$\displaystyle\frac{6}{5*1^2}+\frac{6}{5*2^2}+\frac{6}{5*3^2}+\frac{6}{5*4^2}+\frac{6}{5*5^2}+\frac{6}{5*6^2}+\displaystyle\cdots\cdots=\frac{\pi^2}{5}$
$\displaystyle\frac{6}{6*1^2}+\frac{6}{6*2^2}+\frac{6}{6*3^2}+\frac{6}{6*4^2}+\frac{6}{6*5^2}+\frac{6}{6*6^2}+\displaystyle\cdots\cdots=\frac{\pi^2}{6}$
$\displaystyle\frac{6}{7*1^2}+\frac{6}{7*2^2}+\frac{6}{7*3^2}+\frac{6}{7*4^2}+\frac{6}{7*5^2}+\frac{6}{7*6^2}+\displaystyle\cdots\cdots=\frac{\pi^2}{7}$
$\displaystyle\frac{6}{8*1^2}+\frac{6}{8*2^2}+\frac{6}{8*3^2}+\frac{6}{8*4^2}+\frac{6}{8*5^2}+\frac{6}{8*6^2}+\displaystyle\cdots\cdots=\frac{\pi^2}{8}$
$\displaystyle\frac{6}{9*1^2}+\frac{6}{9*2^2}+\frac{6}{9*3^2}+\frac{6}{9*4^2}+\frac{6}{9*5^2}+\frac{6}{9*6^2}+\displaystyle\cdots\cdots=\frac{\pi^2}{9}$

$\displaystyle\cdots\cdots$

$\small\displaystyle\sum_{k=1}^\infty\frac{6}{a*k^2}=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{6}{(a+1)^nk^2}=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac{6}{(a+1)^{n-1}k^2}=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac{6(a-1)}{a^nk^2}=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac{12(1-a)}{(-1)^ka^nk^2}$

$\small\displaystyle\sum_{k=1}^\infty\frac{6}{a*k^2}=\sum_{k=1}^\infty\frac{1}{k^2}\sum_{n=0}^\infty\frac{6}{(a+1)^{n+1}}=\sum_{k=1}^\infty\frac{1}{k^2}\sum_{n=1}^\infty\frac{6}{(a+1)^n}=\sum_{k=1}^\infty\frac{1}{k^2}\sum_{n=2}^\infty\frac{6}{(a+1)^{n-1}}$

$\small\displaystyle\sum_{k=1}^\infty\frac{6}{a*k^2}=\sum_{k=1}^\infty\frac{(-1)^{k}}{k^2}\sum_{n=0}^\infty\frac{-12}{(a+1)^{n+1}}=\sum_{k=1}^\infty\frac{(-1)^{k}}{k^2}\sum_{n=1}^\infty\frac{-12}{(a+1)^n}=\sum_{k=1}^\infty\frac{(-1)^{k}}{k^2}\sum_{n=2}^\infty\frac{-12}{(a+1)^{n-1}}$

$\small\displaystyle\sum_{k=1}^\infty\frac{6}{a*k^2}=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^2}\sum_{n=0}^\infty\frac{12}{(a+1)^{n+1}}=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^2}\sum_{n=1}^\infty\frac{12}{(a+1)^n}=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^2}\sum_{n=2}^\infty\frac{12}{(a+1)^{n-1}}$

王守恩

谢谢 elim！借此宝地，出一道题(答案是有理数)。
$题:  求和\ \displaystyle\sum_{k=1}^\infty\frac{1}{a*k+k^2}\ \ 其中\ a=1,2,3,4,\cdots\cdots$

elim

王守恩 发表于 2024-1-8 19:36
谢谢 elim！借此宝地，出一道题(答案是有理数)。
\(题:  求和\ \displaystyle\sum_{k=1}^\infty\frac{1}{a ...

题： 求 $\small\displaystyle\sum_{k=1}^\infty\frac{1}{ak+k^2}\;(a\in\mathbb{N}^+)$
$\because\;\small\dfrac{1}{ak+k^2}=\dfrac{1}{a}\big(\dfrac{1}{k}-\dfrac{1}{k+a}\big)$
$\therefore\small\displaystyle\sum_{k=1}^\infty\frac{1}{ak+k^2}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{ak+k^2}=\frac{1}{a}\lim_{n\to\infty}\big(\sum_{k=1}^a\frac{1}{k}-\sum_{k=1}^a\frac{1}{n+k}\big)$
$\qquad\qquad\quad\;\;\small\displaystyle=\frac{1}{a}\sum_{k=1}^a\frac{1}{k}=\frac{1}{a}\big(1+\frac{1}{2}+\cdots+\frac{1}{a}\big)$

回到主题：期待反馈

题：求和 $\displaystyle\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty\frac{1}{k2^n+1}.$
所论二重级数绝对收敛(被优级数$\displaystyle\sum_{k=1}^{\infty}\frac1{k^2}\sum_{n=0}^{\infty}\frac1{2^n}$.控制)因而可重排：
$\small\displaystyle\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k(k2^n+1)}.$
令$f_n(x)=\small\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k-1}x^{k2^n+1}}{k(k2^n+1)}\;(x\in[0,1],\;f(x)=\displaystyle\sum_{n=0}^{\infty}f_n(x)).$
由 $f_n'(x)=\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k-1}(x^{2^n})^k}k=\ln(1+x^{2^n}).$ 知,
$\displaystyle f'(x)=\sum_{n=0}^{\infty}\ln(1+x^{2^n})=\ln\left(\prod_{n=0}^{\infty}(1+x^{2^n})\right)$
$\qquad\;\;=\ln(1+x+x^2+x^3+\cdots)=-\ln(1-x)$
可见$f(1)=\small\displaystyle -\int_0^1\ln(1-x)\,dx=1.$

王守恩

谢谢 elim！$精彩！\displaystyle\frac{1}{ak+k^2}=\frac{1}{a}\bigg(\frac{1}{k}-\frac{1}{k+a}\bigg).\ \  精彩！$谢谢 elim！

$\small\displaystyle\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty\frac{1}{2^n*k+1}=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{(-1)^{k-1}}{2^n*k^2+k}=\bigg(\sum_{k=1}^\infty\frac{1}{k^2+k}\bigg)\frac{2^{\infty}-1}{2^{\infty}}=\sum_{k=1}^\infty\frac{1}{k^2+k}=1$

最后一步需要补充"精彩",  就好了。

$可以拓展:  求和\ \displaystyle\sum_{k=1}^\infty\frac{1}{a*k+k^2}.\ \ 其中\ a\ 可以是正整数(包括主题"1"),可以是正数,也可以是负数。$

王守恩

谢谢 elim！继续讨教。

$求证:\small\displaystyle\sum_{n=0}^\infty\sum_{k=1}^\infty\frac{1}{a^nk^2}=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{1}{a^nk^2}=\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{a}{a^nk^2}=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{a}{a^nk^2}=\sum_{k=1}^\infty\frac{a}{(a-1)k^2}=\frac{a\pi^2}{6(a-1)}$

elim

提示 $\displaystyle\sum_{n=0}^\infty\sum_{k=1}^\infty\frac{1}{a^nk^2}=\big(\sum_{n=0}^\infty\frac{1}{a^n}\big)\big(\sum_{k=1}^\infty\frac{1}{k^2}\big)$

回到主题：期待反馈

题：求和 $\displaystyle\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\sum_{n=0}^\infty\frac{1}{k2^n+1}.$
所论二重级数绝对收敛(被优级数$\displaystyle\sum_{k=1}^{\infty}\frac1{k^2}\sum_{n=0}^{\infty}\frac1{2^n}$.控制)因而可重排：
$\small\displaystyle\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k(k2^n+1)}.$
令$f_n(x)=\small\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k-1}x^{k2^n+1}}{k(k2^n+1)}\;(x\in[0,1],\;f(x)=\displaystyle\sum_{n=0}^{\infty}f_n(x)).$
由 $f_n'(x)=\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k-1}(x^{2^n})^k}k=\ln(1+x^{2^n}).$ 知,
$\displaystyle f'(x)=\sum_{n=0}^{\infty}\ln(1+x^{2^n})=\ln\left(\prod_{n=0}^{\infty}(1+x^{2^n})\right)$
$\qquad\;\;=\ln(1+x+x^2+x^3+\cdots)=-\ln(1-x)$
可见$f(1)=\small\displaystyle -\int_0^1\ln(1-x)\,dx=1.$