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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 $b_k$ thats 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.

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Do you mean the sum over $\frac{5 \sin^2 k} {k!}$ ? – Stefan Oct 16 '12 at 19:31
$\sin^2 k\le 1$ for all $k$ and the series for $e$ converges so... – anon Oct 16 '12 at 19:36
Yeep.Corrections made – Koba Oct 16 '12 at 20:01
up vote 8 down vote accepted

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?

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

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