0
$\begingroup$

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

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

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.

$\endgroup$
0
$\begingroup$

Hint

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.

I am sure that you can take from here.

$\endgroup$
0
$\begingroup$

It converges absolutly due to the estimate $$|\cos^{2}(n)|\leq 1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.