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I have been trying to show the convergence of this series but I can't seem to find a way to do it. $$\sum_{k=1}^\infty \frac{1}{k}\cdot \sin\frac{(-1)^k}{1+k^2}$$

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Using the limit comparison test for the following:

$$a_k=\frac{\sin\frac1{k^2+1}}k\;\;,\;\;\;b_k:=\frac1{k^2+1}\implies \frac{a_k}{b_k}=\frac1k\frac{\sin\frac1{k^2+1}}{\frac1{k^2+1}}\xrightarrow[k\to\infty]{}0\cdot1=0$$

so $\;\sum a_k\;$ converges because $\;\sum b_k\;$ does, and thus your series converges absolutely ( since

$$\;\left|\sin\cfrac{(-1)^k}{k^2+1}\right|=\sin\cfrac1{k^2+1}\;)$$

and then it converges.

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Since for $k\geq 1$, $$\left|\frac{1}{k}\cdot \sin\left(\frac{(-1)^k}{1+k^2}\right)\right|=\frac{1}{k}\cdot \sin\left(\frac{1}{1+k^2}\right)\sim \frac{1}{k}\cdot\frac{1}{1+k^2}\sim \frac{1}{k^3}$$ and $3>1$, then the series converges absolutely and therefore it is also convergent.

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  • $\begingroup$ As usual, asymptotic comparison yields the quickest solution. $\endgroup$ – Gabriel Romon Nov 14 '16 at 13:34
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    $\begingroup$ In fact you don't even need to asymptotically compare: $$\left\lvert \sin\left(\frac{(-1)^k}{1+k^2}\right) \right\rvert \leq \left\lvert \frac{(-1)^k}{1+k^2} \right\rvert$$ with an actual inequality for all $k\geq1$. $\endgroup$ – tomsmeding Nov 14 '16 at 17:16
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Note that $\sin((-1)^k/(1+k^2))=\sin(1/(1+k^2))$ if $k$ is even and $-\sin(1/(1+k^2))$ is $k$ is odd. Further, as $k\to\infty$ we have $(1/k)\sin(1/(1+k^2))\downarrow0$. So you are in the shape of $\sum_na_n$ where $a_n$ are alternating, $a_n\to0$. Thus by Leibnitz test the series converges.

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  • $\begingroup$ To use Dirichlet's series you need a series $\;\sum a_kb_k\;$ , with $\;a_k\;$ non-negative and monotonically descending to zero and the partial sums of $\;\sum b_k\;$ bounded. What you seem to use is Leibniz Test or alternating series test. $\endgroup$ – DonAntonio Nov 14 '16 at 12:04
  • $\begingroup$ You may be right. I have forgotten the exact name as I did thee more than a year back. Editing it. $\endgroup$ – Landon Carter Nov 14 '16 at 12:06
  • $\begingroup$ Note that this argument shows that the series converges at least conditionally. But it doesn't show that the series converges absolutely (which it does). $\endgroup$ – Martin Argerami Nov 14 '16 at 15:02

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