分母に3次式の総和
分母に3次式の総和
次の総和を求めよ。
\[ \sum_{k=1}^{\infty}\frac{1}{\left(4k\right)^{3}-4k}=? \]
次の総和を求めよ。
\[ \sum_{k=1}^{\infty}\frac{1}{\left(4k\right)^{3}-4k}=? \]
\begin{align*}
\sum_{k=1}^{\infty}\frac{1}{\left(4k\right)^{3}-4k} & =\sum_{k=1}^{\infty}\frac{1}{4k\left(\left(4k\right)^{2}-1\right)}\\
& =\sum_{k=1}^{\infty}\frac{1}{4k\left(4k-1\right)\left(4k+1\right)}\\
& =\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{4k}\left(\frac{1}{4k-1}-\frac{1}{4k+1}\right)\\
& =\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{4k-1}-\frac{1}{4k}-\frac{1}{4k}+\frac{1}{4k+1}\right)\\
& =\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{4k-1}-\frac{1}{4k}+\frac{1}{4k+1}-\frac{1}{4k+2}-\frac{1}{4k}+\frac{1}{4k+2}\right)\\
& =\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{4k-1}-\frac{1}{4k}+\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{4k}-\frac{1}{4k+2}\right)\\
& =\frac{1}{2}\sum_{k=3}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\frac{1}{4}\sum_{k=1}^{\infty}\left(\frac{1}{2k}-\frac{1}{2k+1}\right)\\
& =\frac{1}{2}\left(\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\left(1-\frac{1}{2}\right)\right)-\frac{1}{4}\sum_{k=2}^{\infty}\left(\frac{\left(-1\right)^{k}}{k}\right)\\
& =\frac{1}{2}\left(\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\left(1-\frac{1}{2}\right)\right)-\frac{1}{4}\sum_{k=1}^{\infty}\left(\frac{\left(-1\right)^{k}}{k}+1\right)\\
& =\left(\frac{1}{2}+\frac{1}{4}\right)\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\frac{1}{2}\left(1-\frac{1}{2}\right)-\frac{1}{4}\\
& =\frac{3}{4}\log2-\frac{1}{2}
\end{align*}
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