Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – Daniel Fischer Jul 15 '13 at 15:13
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. – AAP Jul 15 '13 at 15:19
It may, but it does not always. I have an example of $a_n$ with finite sum so that the modified series diverges. – Daniel Fischer Jul 15 '13 at 15:21
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. – AAP Jul 15 '13 at 15:23
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? – Mlazhinka Shung Gronzalez LeWy Jul 15 '13 at 15:27
up vote 1 down vote accepted

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 $n\ge 2$.

share|cite|improve this answer
I was just about to add that counterexample. (+1) – robjohn Jul 15 '13 at 15:33
@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. :) – 23rd Jul 15 '13 at 15:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.