有回味的题目

王守恩

\(\displaystyle\sum_{k=1}^{\infty}\frac{1}{4k^2\sqrt{4k^2-1}}\)

\(\displaystyle<\sum_{k=1}^{\infty}\frac{1}{\sqrt{4k^2}(4k^2-1)}\)

\(\displaystyle=\sum_{k=1}^{\infty}\frac{1}{(2k-1)2k(2k+1)}\)

\(\displaystyle=\frac{1}{2}*\sum_{k=1}^{\infty}\frac{(2k+1)-(2k-1)}{(2k-1)2k(2k+1)}\)

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

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

\(\displaystyle=\frac{1}{2}*\sum_{k=1}^{\infty}\bigg(\big(\frac{1}{2k-1}-\frac{1}{2k}\big)-\big(\frac{1}{2k}-\frac{1}{2k+1}\big)\bigg)\)

\(\displaystyle=\frac{1}{2}*\bigg(\ln2-\big(1-\ln2\big)\bigg)\)

\(\displaystyle=\ln2-\frac{1}{2}\)