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How would you determine the character of this series?

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

I have been studying it for 2 days and no result! I tried some things, like since the $\sin(n)$ changes its sign, I started studying the sequence of the absolute value of $a_n$, and since all terms are strictly positive minus $\sin(n)$, $|a_n|$ will be a non-negative one.

So I studied the limit of $|a_n|$ and I found that this is zero, so it can converge. Now I should prove the convergence of $\sum|a_n|$ to prove the convergence of $\sum a_n$ but I don't find a way to that. I tried every method I know, the one that I think could be the right is:

$\frac{(n+2)|\sin(n)|}{n^2(\ln(1+\sqrt n))^2} < \frac{(n+2)}{n^2(\ln(1+\sqrt n))^2}$ because $0<|\sin(n)|<1$. From here I tried Cauchy criteria because with the asymptotic comparison I found nothing. And so with Cauchy one (I refer to the $\sum_{n=1}^{\infty} a_n$ = $\sum_{n=1}^{+\infty} 2^na_{2^n}$)

Can anyone help me?

Anne Bauval
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Gabriele Zinnari
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2 Answers2

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Your series is absolutely convergent, since $$\forall n\ge2,\quad\frac{(n+2)|\sin n|}{n^2\ln^2(1+\sqrt n)}\le\frac{2n}{n^2\ln^2(\sqrt n)}=\frac8{n\ln^2n}$$ and $\sum_{n=2}^\infty\frac1{n\ln^2n}$ converges (it is an instance of Bertrand series).

Anne Bauval
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  • Thank you for the help both with editing my question and for providing a good answer. I choose the other one just because it is crystal clear how i shoul had solved it, but your answer is great too. – Gabriele Zinnari Sep 20 '23 at 07:40
  • Mine is theoretically simpler (upper bound instead of equivalent), so I am not sure you should have solved it otherwise, and the links I provided have the advantage of giving various proofs for the convergence of $\sum_{n=2}^\infty\frac1{n\ln^2n}$, instead of choosing one. – Anne Bauval Sep 20 '23 at 10:19
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Do you know the asymptotic criterions ? If $(u_n)$ and $(v_n)$ are non-negative sequences such that $u_n \sim v_n$, then $\sum u_n$ and $\sum v_n$ have the same nature.

Here, for all $n$, $$ \left|\frac{(n + 2)\sin(n)}{n^2\ln(1 + \sqrt{n})^2}\right| \leqslant \left|\frac{n + 2}{n^2\ln(1 + \sqrt{n})^2}\right| \sim \frac{1}{n\ln(\sqrt{n})^2} = \frac{4}{n\ln(n)^2}. $$ Therefore, all you have to prove is the convergence of $\sum \frac{1}{n\ln(n)^2}$. $f : x \mapsto \frac{1}{x\ln(x)^2}$ decreases on $[2,+\infty[$ so we can use the series-integral criterion and, $$ \int f(x) \, dx = \int \frac{1}{x\ln(x)^2} \, dx = \int \frac{\ln'(x)}{\ln(x)^2} \, dx = -\frac{1}{\ln(x)} + \textrm{cste}, $$ which converges when $x \rightarrow +\infty$.

Cactus
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