一道级数求和

ZhuYue286

请问下式有无较"帅气"的裂项方法证明:
\[\sum_{k=1}^{\infty}{\frac{1}{2k\left( 2k+1 \right)}}-\frac{1}{4}=\sum_{k=1}^{\infty}{\frac{1}{2k\left( 2k+1 \right) \left( 2k+2 \right)}}\]

睡神

\begin{align*}
\sum_{k=1}^{\infty}{\frac{1}{2k\left( 2k+1 \right) \left( 2k+2 \right)}} & = \sum_{k=1}^{\infty}{\dfrac{1}{2k\left( 2k+1 \right)}}-\sum_{k=1}^{\infty}{\dfrac{1}{2k\left( 2k+2 \right)}}\\
& = \sum_{k=1}^{\infty}{\dfrac{1}{2k\left( 2k+1 \right)}}-\dfrac{1}{2}\sum_{k=1}^{\infty}{\left(\dfrac{1}{2k}-\dfrac{1}{2k+2}\right)}\\
& =\sum_{k=1}^{\infty}{\frac{1}{2k\left( 2k+1 \right)}}-\frac{1}{4}
\end{align*}
酱紫？