Use comparison test to determine whether the series converges or not. $$\sum_{k=1}^{\infty} \frac{5\sin^2 k}{k!}$$

Attempt: My guess is that it converges. The problem I am having is that I don't know what to compare it with. I am trying to find series $b_k$ that's is greater. For example, $\frac{1}{k}$ is greater for $k>3$ or something like that. But that's harmonic series that diverge. So I am not quite sure what to compare it with. Hints please.

  • 4
    $\begingroup$ Do you mean the sum over $\frac{5 \sin^2 k} {k!}$ ? $\endgroup$ – Stefan Oct 16 '12 at 19:31
  • 1
    $\begingroup$ $\sin^2 k\le 1$ for all $k$ and the series for $e$ converges so... $\endgroup$ – anon Oct 16 '12 at 19:36

I assume that the series is $$\sum_{k=1}^{\infty} \frac{5\sin^2(k)}{k!}.$$

Hint 1: What do you know about $a$ and $b$ in $a\leq \sin^2(k) \leq b$?

Hint 2 : What can you say of $\frac{5\sin^2(k)}{k!}$ compared to $\frac{5b}{k!}$

Can you conclude from here?

  • $\begingroup$ Well, $0\leq \sin^2k \leq 1$. Ok I got you thanks. $\endgroup$ – Koba Oct 16 '12 at 19:50

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.