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Test the convergence of the following series:

$$\sum\frac{1}{n^2(1+\frac{1}{2}sin\frac{n\pi}{4})}$$


I have tried by ratio test,i.e.,$lim\frac{a_{n+1}}{a_n}=l$,then $\sum u_n$ will be convergent if $l<1$.But nothing can't be said from the form of the ratio I am getting.May be,my approach is wrong.

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Note that since $-1\le \sin(n\pi/4)\le 1$ for all $n$, then

$$\frac23\le \frac{1}{1+\frac12 \sin(n\pi/4)}\le 2$$

Therefore, we find that

$$\frac23 \sum_{n=1}^N \frac1{n^2}\le \sum_{n=1}^N \frac1{n^2\left(1+\frac12 \sin(n\pi/4)\right)}\le 2 \sum_{n=1}^N \frac1{n^2}$$

Can you finish now?

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  • $\begingroup$ Yes,I can.Thank you :) $\endgroup$ – P.B. Jun 4 '16 at 15:22
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    $\begingroup$ Pleased to hear! And you're quite welcome. My pleasure. -Mark $\endgroup$ – Mark Viola Jun 4 '16 at 15:25
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hint:$$\sum\frac{1}{n^2(1+\frac{1}{2}\sin\frac{n\pi}{4})}\leq 2\sum\frac{1}{n^2}$$

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    $\begingroup$ That inequality does not hold. $\endgroup$ – Mark Viola Jun 4 '16 at 15:14

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