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Suppose $\sum_{n=1}^{\infty}{a_n}$ converges, and $a_n > 0$. Does $$\sum_{n=1}^{\infty}{\dfrac{\sin(\sqrt{a_n})}{\sqrt{n}+na_n}}$$ converge or diverge?

Attempt: I was able to prove that it diverges, as shown below, but could not find an example.

Claim: $\sum_{n=1}^{\infty}{\dfrac{1}{\sqrt{n}+na_n}}$ diverges. Proof: Since $\sum_{n=1}^{\infty}{a_n}$ converges, there exists a $n\geq n_0$ such that $$0 \leq a_n \leq 1$$ which gives, $$\dfrac{1}{\sqrt{n}+na_n} \geq \dfrac{1}{\sqrt{n}+n}$$ proving the claim. Doing a limit comparison test for$\sum_{n=1}^{\infty}{\dfrac{\sin(\sqrt{a_n})}{\sqrt{n}+na_n}}$ with $\sum_{n=1}^{\infty}{\dfrac{1}{\sqrt{n}+na_n}}$ we get $$\lim_{n\rightarrow \infty}{\dfrac{\sin(\sqrt{a_n})}{\sqrt{n}+na_n}\cdot \dfrac{\sqrt{n}+na_n}{1}} = \sin(\sqrt{a_n}) < \infty$$ and hence the given series diverges. However, I am having trouble finding an example.

Thanks in advance!

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  • $\begingroup$ But $\sin \sqrt{a_n} \to 0$, that can easily force convergence (consider $a_n = \frac{1}{n^2}$). To see if/that is always the case is not so easy/obvious. $\endgroup$
    – Daniel Fischer
    Jul 15, 2013 at 15:13
  • $\begingroup$ Right, but the series may fail to converge even if $\lim_{n\rightarrow \infty}{b_n}=0$ where $b_n$ is the series in question. $\endgroup$
    – AAP
    Jul 15, 2013 at 15:19
  • $\begingroup$ It may, but it does not always. I have an example of $a_n$ with finite sum so that the modified series diverges. $\endgroup$
    – Daniel Fischer
    Jul 15, 2013 at 15:21
  • $\begingroup$ True. But that is what the question is, does it always fail to converge for the above series? By the reasoning I showed above, it seems so. $\endgroup$
    – AAP
    Jul 15, 2013 at 15:23
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    $\begingroup$ Not always. Notice the comparison test gave you limit zero. In this case you only get one implication about convergence between the series. Which one? $\endgroup$
    – OR.
    Jul 15, 2013 at 15:27

1 Answer 1

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Hint: It is easy to see it converges for some $(a_n)$. To see it could diverge, you may consider $a_n=\frac{1}{n(\log n)^2}$ for $n\ge 2$.

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  • $\begingroup$ I was just about to add that counterexample. (+1) $\endgroup$
    – robjohn ♦
    Jul 15, 2013 at 15:33
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    $\begingroup$ @robjohn: Thank you. I'm using iPad and it is quite inconvenient for typing, so I decided to only give a hint rather than a full answer. Probably that's why I could post the answer earlier than you. :) $\endgroup$
    – 23rd
    Jul 15, 2013 at 15:39

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