【微分積分】1-2-2 級数の収束と発散｜問題集

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

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

\(\displaystyle a_{n+1}^2=1-\frac{2a_{n+1}}{a_n}\)
\(\displaystyle a_{n+1}^2+\frac{2a_{n+1}}{a_n}-1=0\)
\(a_n=\tan \theta_n\)、\(\displaystyle \theta_n=\frac{\pi}{2^{n+1}}\)とおくと、
\(\displaystyle \tan^2\theta_{n+1}+\frac{2\tan\theta_{n+1}}{\tan\theta_n}-1=0\)
また、\(\theta_n=2\theta_{n+1}\)なので、\(2\)倍角の公式より、
\(\displaystyle \tan2\theta_{n+1}=\frac{2\tan\theta_{n+1}}{1-\tan^2\theta_{n+1}}\)
\(\displaystyle \tan\theta_n=\frac{2\tan\theta_{n+1}}{1-\tan^2\theta_{n+1}}\)
\(\displaystyle 1-\tan^2\theta_{n+1}=\frac{2\tan\theta_{n+1}}{\tan\theta_n}\)
\(\displaystyle \tan^2\theta_{n+1}+\frac{2\tan\theta_{n+1}}{\tan\theta_n}-1=0\)
よって、
\(\displaystyle a_{n+1}=\frac{1}{a_{n+1}}-\frac{2}{a_n}\)が成り立つ。

\(\displaystyle \theta_n=\frac{\pi}{2^{n+1}}\)とおくと、
\(\displaystyle \tan\theta_n=\frac{1}{\tan\theta_n}-\frac{2}{\tan2\theta_n}\)
\(\displaystyle \frac{1}{2^n}\tan\theta_n=\frac{1}{2^n}・\frac{1}{\tan\theta_n}-\frac{1}{2^{n-1}}・\frac{1}{\tan2\theta_n}\)
すなわち、
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{\pi}{2^{n+1}}\)
\(\displaystyle =\sum_{n=1}^{\infty}\left(\frac{1}{2^n}・\frac{1}{\tan\theta_n}-\frac{1}{2^{n-1}}・\frac{1}{\tan2\theta_n}\right)\)
\(\displaystyle =\frac{1}{2^n}・\frac{1}{\tan\theta_n}-\frac{1}{2^0}・\frac{1}{\tan2\theta_0}\)
\(\displaystyle =\frac{1}{2^n}・\frac{1}{\tan\theta_n}\)
\(\displaystyle =\frac{1}{2^n}・\frac{1}{\tan\frac{\pi}{2^{n+1}}}\)
\(\displaystyle =\frac{2}{\pi}\left(\frac{\pi}{2^{n+1}}・\frac{1}{\tan\frac{\pi}{2^{n+1}}}\right)\)
よって、
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{\pi}{2^{n+1}}\)
\(\displaystyle =\lim_{n\to\infty}\frac{2}{\pi}\left(\frac{\pi}{2^{n+1}}・\frac{1}{\tan\frac{\pi}{2^{n+1}}}\right)\)
小角近似\(\displaystyle \lim_{n\to\infty}\frac{\pi}{2^{n+1}}・\frac{1}{\tan\frac{\pi}{2^{n+1}}}=1\)より、
\(\displaystyle =\frac{2}{\pi}\)