【微分積分】5-4-1 収束半径｜問題集
1.次の整級数の収束半径を求めなさい。
(1)\(\displaystyle \sum_{n=0}^{\infty}\frac{1}{2n+1}x^{2n+1}\)
\(\displaystyle a_n=\frac{1}{2n+1}\)とおくと、
\(\displaystyle \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)
\(\displaystyle =\lim_{n\to\infty}\frac{1}{2n+1}・\frac{2n+3}{1}\)
\(\displaystyle =\lim_{n\to\infty}\frac{2n+3}{2n+1}\)
\(\displaystyle =1\)
よって、収束半径は\(1\)
\(\displaystyle \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)
\(\displaystyle =\lim_{n\to\infty}\frac{1}{2n+1}・\frac{2n+3}{1}\)
\(\displaystyle =\lim_{n\to\infty}\frac{2n+3}{2n+1}\)
\(\displaystyle =1\)
よって、収束半径は\(1\)
(2)\(\displaystyle \sum_{n=0}^{\infty}\frac{2^{2n}}{2n!}x^{2n}\)
\(\displaystyle a_n=\frac{2^{2n}}{2n!}\)とおくと、
\(\displaystyle \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)
\(\displaystyle =\lim_{n\to\infty}\frac{2^{2n}}{2n!}・\frac{(2n+2)!}{2^{2n+2}}\)
\(\displaystyle =\lim_{n\to\infty}\frac{(2n+2)(2n+1)}{4}\)
\(\displaystyle =\infty\)
よって、収束半径は\(\infty\)
\(\displaystyle \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)
\(\displaystyle =\lim_{n\to\infty}\frac{2^{2n}}{2n!}・\frac{(2n+2)!}{2^{2n+2}}\)
\(\displaystyle =\lim_{n\to\infty}\frac{(2n+2)(2n+1)}{4}\)
\(\displaystyle =\infty\)
よって、収束半径は\(\infty\)
(3)\(\displaystyle \sum_{n=0}^{\infty}n!x^n\)
\(\displaystyle a_n=n!\)とおくと、
\(\displaystyle \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)
\(\displaystyle =\lim_{n\to\infty}\frac{n!}{(n+1)!}\)
\(\displaystyle =\lim_{n\to\infty}\frac{1}{n+1}\)
\(\displaystyle =0\)
よって、収束半径は\(0\)
\(\displaystyle \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)
\(\displaystyle =\lim_{n\to\infty}\frac{n!}{(n+1)!}\)
\(\displaystyle =\lim_{n\to\infty}\frac{1}{n+1}\)
\(\displaystyle =0\)
よって、収束半径は\(0\)
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