【微分積分】1-2-1 級数の和｜問題集

\(\displaystyle =\frac{1}{n^2}・\frac{n(n+1)}{2}\)
\(\displaystyle =\frac{n^2+n}{2n^2}\)
\(\displaystyle =\frac{1}{2}+\frac{1}{2n}\)
よって、
\(\displaystyle =\lim_{n\to\infty}\left(\frac{1}{2}+\frac{1}{2n}\right)\)
\(\displaystyle =\frac{1}{2}\)

\(\displaystyle =\frac{1}{n^3}・\frac{n(n+1)(2n+1)}{6}\)
\(\displaystyle =\frac{2n^3+3n^2+n}{6n^3}\)
よって、
\(\displaystyle =\lim_{n\to\infty}\frac{2n^3+3n^2+n}{6n^3}\)
\(\displaystyle =\frac{1}{3}\)

\(\displaystyle =\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(\displaystyle =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)\)
\(\displaystyle \ \ \ +\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(\displaystyle =1-\frac{1}{n+1}\)
よって、
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1\)

\(\displaystyle =\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+3)}\)
\(\displaystyle =\sum_{n=1}^{\infty}\frac{1}{4}\left(\frac{1}{2n-1}-\frac{1}{2n+3}\right)\)
\(\displaystyle =\frac{1}{4}・\left(1-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\cdots\)
\(\displaystyle \ \ +\left(\frac{1}{2n-3}-\frac{1}{2n+1}\right)+\left(\frac{1}{2n-1}-\frac{1}{2n+3}\right)\)
\(\displaystyle =\frac{1}{4}\left(1+\frac{1}{3}-\frac{1}{2n+1}-\frac{1}{2n+3}\right)\)
\(\displaystyle =\frac{1}{3}-\frac{1}{4}\left(\frac{1}{2n+1}+\frac{1}{2n+3}\right)\)
よって、
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n^2+4n-3}\)
\(\displaystyle =\lim_{n\to\infty}\left\{\frac{1}{3}-\frac{1}{4}\left(\frac{1}{2n+1}+\frac{1}{2n+3}\right)\right\}\)
\(\displaystyle =\frac{1}{3}\)