# How do I determine the convergence of this series?

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

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.

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?

• Yes that's the $b_k$ I was looking for, the comparison test then gives c=1. Thanks!! May 6, 2021 at 10:48
• 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? May 6, 2021 at 10:54
• 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. May 6, 2021 at 11:03

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.

• Hi Bernard ! There is a very nice solution for the infinite sum. With ten additions, I get an absolute error of $2.5\times 10^{-8}$. May 6, 2021 at 13:19
• Only 10 terms? I'm no specialist of convergence speed, but that looks like a miracle! May 6, 2021 at 13:44
• 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. May 6, 2021 at 13:55