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Author |
hbghlyj
Post time 2024-2-27 03:37
If $x$ is any complex number not equal to a positive integer, then
$$
\sum_{n=1}^{\infty} \frac{1}{n(n-x)} \sum_{m=1}^{n-1} \frac{1}{m-x}=\sum_{n=1}^{\infty} \frac{1}{n^{2}(n-x)} .
$$
Proof. Fix $x \in \mathbf{C} \backslash \mathbf{Z}^{+}$. Let $S$ denote the left hand side. By partial fractions,
$$
\begin{aligned}
S &=\sum_{n=1}^{\infty} \sum_{m=1}^{n-1}\left(\frac{1}{n(n-m)(m-x)}-\frac{1}{n(n-m)(n-x)}\right) \\
&=\sum_{m=1}^{\infty} \frac{1}{m-x} \sum_{n=m+1}^{\infty} \frac{1}{n(n-m)}-\sum_{n=1}^{\infty} \frac{1}{n(n-x)} \sum_{m=1}^{n-1} \frac{1}{n-m} \\
&=\sum_{m=1}^{\infty} \frac{1}{m(m-x)} \sum_{n=m+1}^{\infty}\left(\frac{1}{n-m}-\frac{1}{n}\right)-\sum_{n=1}^{\infty} \frac{1}{n(n-x)} \sum_{m=1}^{n-1} \frac{1}{m}
\end{aligned}
$$
Now for fixed $m \in \mathbf{Z}^{+}$,
$$
\begin{aligned}
\sum_{n=m+1}^{\infty}\left(\frac{1}{n-m}-\frac{1}{n}\right) &=\lim _{N \rightarrow \infty} \sum_{n=m+1}^{N}\left(\frac{1}{n-m}-\frac{1}{n}\right)=\sum_{n=1}^{m} \frac{1}{n}-\lim _{N \rightarrow \infty} \sum_{n=1}^{m} \frac{1}{N-n+1} \\
&=\sum_{n=1}^{m} \frac{1}{n}
\end{aligned}
$$
since $m$ is fixed. Therefore, we have
$$
\begin{aligned}
S &=\sum_{m=1}^{\infty} \frac{1}{m(m-x)} \sum_{n=1}^{m} \frac{1}{n}-\sum_{n=1}^{\infty} \frac{1}{n(n-x)} \sum_{m=1}^{n-1} \frac{1}{m}=\sum_{n=1}^{\infty} \frac{1}{n(n-x)}\left(\sum_{m=1}^{n} \frac{1}{m}-\sum_{m=1}^{n-1} \frac{1}{m}\right) \\
&=\sum_{n=1}^{\infty} \frac{1}{n^{2}(n-x)}
\end{aligned}
$$
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