证明恒等式
\[ \sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{1}{nm(n+m)} =2\sum_{n=1}^ \infty \frac{1}{n^3} \] 这题，一看我就不会 裂项, \[ \sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{1}{nm(n+m)} = \sum _{n=1}^{\infty } \frac{1}{n^2}(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}) =\sum _{n=1}^{\infty }\sum _{m=1}^{n }\frac{1}{n^2m}=2 \zeta (3) \] 我们设$f(x)=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{x^{n+m}}{nm(n+m)}$
于是$xf'(x)=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{x^{n+m}}{nm}=(\sum_{n=1}^{\infty}\frac{x^n}n)^2=(-\ln(1-x))^2$
于是我们得到对于$0\le x\lt 1$有，$f(x)=\int_0^x \frac{\ln(1-x)^2}{x} dx$, 于是等式左边等于$\int_0^1 \frac{\ln^2(1-x)}{x} dx=\int_0^1 \frac{ln^2(x)}{1-x}dx$
又因为$\frac{ln^2(x)}{1-x}=\sum_{k=0}^{\infty} x^k \ln^2(x)$
我们得到左边=$\sum_{k=0}^{\infty} \int_0^1 x^k \ln^2(x)dx$
其中$\int_0^1 x^k \ln^2(x)dx = \int_{-\infty}^0 \exp((k+1)t) t^2 dt = \int_0^{+\infty}t^2\exp(-(k+1)t)dt = t^2\exp(-t)|_0^{+\infty}+\int_0^{+\infty}\frac{2t}{k+1}\exp(-(k+1)t)dt=\int_0^{+\infty}\frac{2}{(k+1)^2}\exp(-(k+1)t)dt=\frac{2}{(k+1)^3}$
由此得到最终结果。 一直在思考是否可以仅仅通过组合变换，不用积分来求出结果，刚刚发现一个方法。记：
\[\sum_{k=1}^n \frac{1}{k} =H_n\]
一方面：
\[\sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{1}{nm(n+m)} =\sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{n+m-m}{n^2m(n+m)}=\sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{1}{n^2}( \frac{1}{m}- \frac{1}{n+m})=\sum_{n=1}^ \infty \frac{H_n}{n^2} \]
另一方面，令：\(k=n+m\)，则：
\[\sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{1}{nm(n+m)} = \sum _{k=2}^{\infty } \frac{1}{k}(\frac{1}{1(k-1)}+\frac{1}{2(k-2)}+\frac{1}{3(k-3)}+...+\frac{1}{(k-1)1}) =\sum _{k=2}^{\infty } \frac{1}{k} \sum _{i=1}^{k-1 }\frac{1}{i(k-i)}\]
\[=\sum _{k=2}^{\infty } \frac{1}{k^2} \sum _{i=1}^{k-1 }(\frac{1}{i}+\frac{1}{k-i})= 2 \sum _{k=2}^{\infty } \frac{1}{k^2} \sum _{i=1}^{k-1 }\frac{1}{i}= 2 \sum _{k=2}^{\infty } \frac{1}{k^2} (H_k-\frac{1}{k})=2 \sum _{n=1}^{\infty } \frac{1}{n^2} (H_n-\frac{1}{n})=2 \sum _{n=1}^{\infty } \frac{H_n}{n^2} -2 \zeta (3)\]
即：\
即：\[\sum _{n=1}^{\infty } \frac{H_n}{n^2}=2 \zeta (3)\]
本帖最后由 yigo 于 2024-4-2 15:43 编辑
再来求一个.
\[ \sum _{n=1}^{\infty } \frac{H_n}{n^3}=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{n^2m(n+m)}=\frac{1}{2}(\sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{n^2m(n+m)}+\sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{m^2n(n+m)})=\frac{1}{2} \sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{n^2m^2}= \frac{1}{2}(\sum _{n=1}^{\infty }\frac{1}{n^2})^2=\frac{\pi^4}{72}\]
\[ \sum _{n=1}^{\infty } \frac{H_n}{n^4}\]不会算了。
\[ \sum _{n=1}^{\infty } \frac{H_n}{n^5}=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{n^4m(n+m)}=\frac{1}{2}(\sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{n^4m(n+m)}+\sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{m^4n(n+m)})=\frac{1}{2} \sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{n^3+m^3}{n^4m^4(n+m)}\]
\[ =\frac{1}{2} \sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{(n+m)^3-3nm(n+m)}{n^4m^4(n+m)}=\frac{1}{2} \sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{(n+m)^2}{n^4m^4}-\frac{1}{2} \sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{3}{n^3m^3}\]
\[=\frac{1}{2} \sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{n^2+m^2+2nm}{n^4m^4}-\frac{1}{2} \sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{3}{n^3m^3}=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{n^2m^4}-\frac{1}{2} \sum _{n=1}^{\infty }\sum _{m=1}^{\infty } \frac{1}{n^3m^3}=\zeta (2)\zeta (4)-\frac{1}{2}\zeta ^2(3)\]
我分项的那本电子书(Almost)-impossible-integrals.pdf里面有通项公式，但是好像没证明。
\[ \sum _{n=1}^{\infty } \frac{H_n}{n^k}\] 本帖最后由 yigo 于 2024-4-2 17:29 编辑
拓展一下，\[\sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{1}{nm(n+pm)} =\int_0^1 \frac{\ln(1-x)\ln(1-x^p)}{x} dx=?,(其中p>0)\]
软件好像算不出来，可能没有解析解果。
当\(p=2\)时，
\[\sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{1}{nm(n+2m)} =\int_0^1 \frac{\ln(1-x)\ln(1-x^2)}{x} dx=\int_0^1 \frac{\ln(1-x)^2}{x} dx+\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x} dx=2\zeta(3)-\frac{5}{8}\zeta(3)=\frac{11}{8}\zeta(3)\] 本帖最后由 yigo 于 2024-4-3 21:04 编辑
\[\sum _{n=1}^{\infty } \frac{H_n}{n^4}\]求出来了。
\[记：H_n^{(k)}=\sum _{i=1}^{n}\frac{1}{i^k}=\sum _{i=1}^{\infty } (\frac{1}{i^k}-\frac{1}{(i+n)^k})\]
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\[即：A+B=\zeta(2)\zeta(3)\]
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\[即：4A-B+2C=6\zeta(5)\]
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\[即：C+D=\zeta2)\zeta(3)\]
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\[即：\frac{3}{2}B+D=\zeta2)\zeta(3)\]
联立A、B、C、D，
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\[\frac{3}{2}B+D=\zeta2)\zeta(3)\]
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\ 本帖最后由 yigo 于 2024-4-4 13:14 编辑
上面的计算过程比较乱，整理了下思路，发现分子分母通过一直乘\(n+m\)，可以简化计算，下面通过这种方式计算\( \displaystyle \sum _{n=1}^{\infty } \frac{H_n}{n^p}\)。
记：\( \displaystyle H_n^{(p)}=\sum _{i=1}^{n}\frac{1}{i^p}=\sum _{i=1}^{\infty } \left(\frac{1}{i^p}-\frac{1}{(i+n)^p}\right)=\zeta(p)-\sum _{i=1}^{\infty } \frac{1}{(i+n)^p}\)，\( \displaystyle H_n^{(1)}\)简写为\( \displaystyle H_n\)。
先证明一个恒等式:\( \displaystyle \sum _{n=1}^{\infty } \frac{H_n^{(q)}}{n^p}+\sum _{n=1}^{\infty } \frac{H_n^{(p)}}{n^q}=\zeta(p+q)+\zeta(p)\zeta(q)\)，当\(p,q\ge2\)时。
\( \begin{equation} \begin{split}
\sum _{n=1}^{\infty } \frac{H_n^{(q)}}{n^p}&= \sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{n^p}\left( \frac{1}{m^q}- \frac{1}{(n+m)^q}\right) \\
&=\zeta(p)\zeta(q)-\sum _{n=1}^{\infty } \sum _{m=1}^{\infty }\frac{1}{n^p(n+m)^q}\\
令k=n+m\\
&=\zeta(p)\zeta(q)-\sum _{k=2}^{\infty } \sum _{i=1}^{k-1 }\frac{1}{i^pk^q}\\
&=\zeta(p)\zeta(q)-\sum _{k=2}^{\infty }\frac{H_k^{(p)}-k^p}{k^q}\\
&=\zeta(p)\zeta(q)+\zeta(p+q)-\sum _{n=1}^{\infty }\frac{H_n^{(p)}}{n^q},证毕\\
\end{split} \end{equation} \)
\( \begin{equation} \begin{split}
\sum _{n=1}^{\infty } \frac{H_n}{n^p}&= \sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{n^p}\left( \frac{1}{m}- \frac{1}{n+m}\right) \\
&= \sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{n^{p-1}m(n+m)} \\
&= \sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{n+m}{n^{p-1}m(n+m)^2} \\
&= \sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{n^{p-1}(n+m)^2}+\sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{n^{p-2}m(n+m)^2}\\
&= \sum _{n=1}^{\infty }\frac{1}{n^{p-1}}\left(\zeta(2)-H_n^{(2)}\right)+\sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{n+m}{n^{p-2}m(n+m)^3}\\
&=\left(\zeta(2)\zeta(p-1)-\sum _{n=1}^{\infty }\frac{H_n^{(2)}}{n^{p-1}}\right)+\left(\zeta(3)\zeta(p-2)-\sum _{n=1}^{\infty }\frac{H_n^{(3)}}{n^{p-2}}\right)+\sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{n^{p-3}m(n+m)^3}\\
&=\sum _{k=1}^{p-2}\left(\zeta(k+1)\zeta(p-k)-\sum _{n=1}^{\infty }\frac{H_n^{(k+1)}}{n^{p-k}}\right)+\sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{nm(n+m)^{p-1}}\\
&=\frac{1}{2}\sum _{k=1}^{p-2}\left(2\zeta(k+1)\zeta(p-k)-\sum _{n=1}^{\infty }\frac{H_n^{(k+1)}}{n^{p-k}}-\sum _{n=1}^{\infty }\frac{H_n^{(p-k)}}{n^{k+1}}\right)+\sum _{k=2}^{\infty } \sum _{i=1}^{k-1 } \frac{1}{i(k-i)k^{p-1}}\\
&=\frac{1}{2}\sum _{k=1}^{p-2}\left(2\zeta(k+1)\zeta(p-k)-\zeta(k+1)\zeta(p-k)-\zeta(p+1)\right)+2\sum _{n=1}^{\infty } \frac{H_n-\frac{1}{n}}{n^{p}}\\
&=\frac{1}{2}\left(\sum _{k=1}^{p-2}\zeta(k+1)\zeta(p-k)\right)-\frac{p-2}{2}\zeta(p+1)+2\sum _{n=1}^{\infty } \frac{H_n}{n^{p}}-2\zeta(p+1)\\
即：\\
\sum _{n=1}^{\infty } \frac{H_n}{n^p}&=\frac{p+2}{2}\zeta(p+1)-\frac{1}{2}\sum _{k=1}^{p-2}\zeta(k+1)\zeta(p-k)\\
\end{split} \end{equation} \)
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