# Test for infinite series convergence or divergence

Hi what would be a good test to find convergence or divergence please?

$\sum _1 ^{\infty} (e^kcos^2k)/ \pi ^k$

My attempt I got that converges thanks

• It looks possible since the terms go to $0$ as $k$ goes to infinity. What tests are you familiar with? – abiessu Sep 28 '14 at 13:49
• The partial sum is increasing and bounded, hence it converges. – Frédéric Sep 28 '14 at 13:53

Since $$0\le \cos^2(\hbox{anything})\le1$$ we have $$0\le\frac{e^k\cos^2k}{\pi^k}\le\frac{e^k}{\pi^k}=\Bigl(\frac{e}{\pi}\Bigr)^k\ .$$ And $$\sum_{k=1}^\infty \Bigl(\frac{e}{\pi}\Bigr)^k$$ converges since it is a GP with ratio $e/\pi<1$, so your series converges by the comparison test.
Consider $$I=\sum\limits_{k=1}^{\infty}\dfrac{e^{k}\cos^{2}(k)}{\pi^{k}}$$ $$J=\sum\limits_{k=1}^{\infty}\dfrac{e^{k}\sin^{2}(k)}{\pi^{k}}$$ So $$I+J=\sum\limits_{k=1}^{\infty}\dfrac{e^{k}}{\pi^{k}}=\frac{e}{\pi-e }$$ which is a geometric series.
It converges absolutly due to the estimate $$|\cos^{2}(n)|\leq 1$$