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$$\sum_{k=1}^{+\infty}\sqrt{k}\sin(\frac{1}{k^2})$$

I attempted the problem using the following limit comparison test (I don't know the name):

If $\sum a_k$ and $\sum b_k$ positive series such that when $k\to+\infty$ $$\frac{a_k}{b_k}\to c $$ where $c$ is finite and greater than zero, then $$\sum a_k \text{ converges} \iff \sum b_k \text{converges}$$

I compared with the sum $\sum_{k=1}^{+\infty}\sqrt k$ which diverges, so we get $$\frac{\sqrt{k}\sin(\frac{1}{k^2})}{\sqrt k}=\sin(\frac{1}{k^2})\to0\ as\ k\to+\infty$$ Therefore $$\sum_{k=1}^{+\infty}\sqrt{k}\sin(\frac{1}{k^2})\text{ diverges.}$$

The series actually converges and I believe my mistake is that I allowed $c=0$ in the comparison test. What other method should I use?

After responses, the solution:

Using $b_k=\frac{1}{k^{3/2}}$ the comparison test gives $c=1$, and as $\sum b_k$ converges, the series $$\sum_{k=1}^{+\infty}\sqrt{k}\sin(\frac{1}{k^2})$$ also converges.

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2 Answers 2

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Yes, you made a mistake: in the comparison test, the equivalence $$\sum a_k \text{ converges} \iff \sum b_k \text{ converges}$$ works when $\lim_{k\to \infty}\frac{a_k}{b_k}=c\not=0$. Therefore you can't conclude that the series is divergent.

Hint. Since $\sin(x)=x+o(x)$ as $x\to 0$, consider $b_k=k^{1/2}(1/k)^2=\frac{1}{k^{3/2}}$ and apply the comparison test. What is $c$? Is the series convergent or not?

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  • $\begingroup$ Yes that's the $b_k$ I was looking for, the comparison test then gives c=1. Thanks!! $\endgroup$
    – Geeniee
    May 6, 2021 at 10:48
  • $\begingroup$ I was wondering though, is there a way to determine which series to use as $b_k$ and $a_k$? Considering the choice ''decides'' whether the limit of the quotient diverges or converges? $\endgroup$
    – Geeniee
    May 6, 2021 at 10:54
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    $\begingroup$ In this case you should use the Taylor expansion of $\sin(x)$ at $x_0=0$: $\sin(x)=x+o(x)$. This approach generally works when the term $a_k$ involves "good" infinitesimals. $\endgroup$
    – Robert Z
    May 6, 2021 at 11:03
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Use asymptotic equivalence:

We know that $\sin x\sim_0 x$, therefore, as $\frac 1{k^2}\to 0$ as $k\to\infty$, $$\sqrt k\,\sin\frac1{k^2}\sim_\infty\sqrt k\,\frac 1{k^2}=\frac1{k^{3/2}},$$ a convergent $p$-series.

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  • $\begingroup$ Hi Bernard ! There is a very nice solution for the infinite sum. With ten additions, I get an absolute error of $2.5\times 10^{-8}$. $\endgroup$ May 6, 2021 at 13:19
  • $\begingroup$ Only 10 terms? I'm no specialist of convergence speed, but that looks like a miracle! $\endgroup$
    – Bernard
    May 6, 2021 at 13:44
  • $\begingroup$ Not at all. The first term is already $2.61238$; then it is the summation of $\zeta$ function. This is just Taylor expansion and reversing the order of the summations. It is a pitty that the question did not come. $\endgroup$ May 6, 2021 at 13:55

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