【微分積分】5-4-2 整級数の計算｜問題集

\(\displaystyle a_n=\frac{1}{2^nn}\)とおくと、
\(\displaystyle \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)
\(\displaystyle =\lim_{n\to\infty}\frac{2^{n+1}(n+1)}{2^nn}\)
\(\displaystyle =\lim_{n\to\infty}2\left(1+\frac{1}{n}\right)\)
\(\displaystyle =2\)
\(\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{1}{2^nn}x^n\)とおくと、\(|x|<2\)で項別微分可能。
\(\displaystyle f'(x)=\sum_{n=1}^{\infty}\frac{1}{2^n}x^{n-1}\)
\(\displaystyle =\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n\)
\(\displaystyle =\frac{1}{2}・\frac{1}{1-\frac{x}{2}}\)
\(\displaystyle =\frac{1}{2-x}\)
すなわち、
\(\displaystyle f(x)=-\log(2-x)+C\)
\(f(0)=0\)より、
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^nn}=f(1)=\log2\)

\(\displaystyle \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\)を項別微分すると
\(\displaystyle \sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}\)
よって、
\(\displaystyle \sum_{n=1}^{\infty}\frac{n}{2^n}\)
\(\displaystyle =\frac{\frac{1}{2}}{(1-\frac{1}{2})^2}\)
\(\displaystyle =2\)

\(\displaystyle \sum_{n=1}^{\infty}(3n-1)x^n\)
\(\displaystyle =3\sum_{n=1}^{\infty}nx^n-\sum_{n=1}^{\infty}x^n\)
\(\displaystyle =\frac{3x}{(1-x)^2}-\frac{x}{1-x}\)
\(\displaystyle =\frac{x(2+x)}{(1-x)^2}\)
よって、
\(\displaystyle \sum_{n=1}^{\infty}\frac{3n-1}{3^n}\)
\(\displaystyle =\frac{\frac{1}{3}(2+\frac{1}{3})}{(1-\frac{1}{3})^2}\)
\(\displaystyle =\frac{7}{4}\)

\(\displaystyle \sum_{n=1}^{\infty}n^2x^{n-1}\)
\(\displaystyle =\frac{(1-x)^2-x・2(1-x)・(-1)}{(1-x)^4}\)
\(\displaystyle =\frac{1+x}{(1-x)^3}\)
\(\displaystyle \sum_{n=1}^{\infty}n^2x^n=\frac{x(1+x)}{(1-x)^3}\)
よって、
\(\displaystyle \sum_{n=1}^{\infty}\frac{n^2}{2^n}\)
\(\displaystyle =\frac{\frac{1}{2}(1+\frac{1}{2})}{(1-\frac{1}{2})^3}\)
\(\displaystyle =6\)