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<math>\frac{1}{n+1} < \sum\limits_{i=n+1}^{\infty} \frac{n!}{i!} < \sum\limits_{i=1}^{\infty} \frac{1}{(n+1)^i} = \frac{1}{n} </math>, | <math>\frac{1}{n+1} < \sum\limits_{i=n+1}^{\infty} \frac{n!}{i!} < \sum\limits_{i=1}^{\infty} \frac{1}{(n+1)^i} = \frac{1}{n} </math>, | ||
| − | so for <math>n \geq 2, |\sin(\frac{ | + | so for <math>n \geq 2, |\sin(\frac{\pi}{n+1})| < |\sin(n!e\pi)| < |\sin(\frac{\pi}{n})| </math>. By the comparison test, <math>\sum\limits_{n=1}^{\infty} |\sin(n!e\pi)|^\alpha</math> converges if <math>\sum\limits_{n=1}^{\infty} |\sin(\frac{1}{n})|^\alpha</math> converges, and diverges if <math>\sum\limits_{n=1}^{\infty} |\sin(\frac{1}{n+1})|^\alpha</math> diverges. |
By elementary calculus (namely, the limit comparison test with the harmonic series), absolute convergence occurs if and only if <math>\alpha > 1</math>. | By elementary calculus (namely, the limit comparison test with the harmonic series), absolute convergence occurs if and only if <math>\alpha > 1</math>. | ||
-pw | -pw | ||
Latest revision as of 14:27, 22 July 2008
We have that
$ |\sin(n!e\pi)| = |\sin(n!\pi(\sum\limits_{i=0}^{\infty} \frac{1}{i!}))| = |sin(k\pi + \pi\sum\limits_{i=n+1}^{\infty} \frac{1}{i!})| = |sin(\pi\sum\limits_{i=n+1}^{\infty} \frac{n!}{(i)!})| $, where $ k\in\mathbb{Z} $.
Furthermore, $ \sin(x) $ is monotone on $ [0, \frac{\pi}{2}] $, and
$ \frac{1}{n+1} < \sum\limits_{i=n+1}^{\infty} \frac{n!}{i!} < \sum\limits_{i=1}^{\infty} \frac{1}{(n+1)^i} = \frac{1}{n} $,
so for $ n \geq 2, |\sin(\frac{\pi}{n+1})| < |\sin(n!e\pi)| < |\sin(\frac{\pi}{n})| $. By the comparison test, $ \sum\limits_{n=1}^{\infty} |\sin(n!e\pi)|^\alpha $ converges if $ \sum\limits_{n=1}^{\infty} |\sin(\frac{1}{n})|^\alpha $ converges, and diverges if $ \sum\limits_{n=1}^{\infty} |\sin(\frac{1}{n+1})|^\alpha $ diverges.
By elementary calculus (namely, the limit comparison test with the harmonic series), absolute convergence occurs if and only if $ \alpha > 1 $.
-pw
