I have been trying to show the convergence of this series but I can't seem to find a way to do it. $$\sum_{k=1}^\infty \frac{1}{k}\cdot \sin\frac{(1)^k}{1+k^2}$$
3 Answers
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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As usual, asymptotic comparison yields the quickest solution. – Gabriel Romon Nov 14 '16 at 13:34

5In 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 $k\geq1$. – tomsmeding Nov 14 '16 at 17:16
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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To use Dirichlet's series you need a series $\;\sum a_kb_k\;$ , with $\;a_k\;$ nonnegative 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. – DonAntonio Nov 14 '16 at 12:04

You may be right. I have forgotten the exact name as I did thee more than a year back. Editing it. – Landon Carter Nov 14 '16 at 12:06

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). – Martin Argerami Nov 14 '16 at 15:02