# Convergence/Divergence of $\sum_{k=1}^{\infty}a_k$, $a_k=\frac{1}{k}\sin(\frac{1}{k})$

I'm trying to use the limit comparison test to test for convergence but I've got different results depending on what I use as my $b_k$

If I use $b_k=\frac{1}{k^{2}}$, I have $\frac{a_k}{b_k}\to 1$ $\sum_{k=1}^{\infty}b_k$ converges, so $\sum_{k=1}^{\infty}a_k$ will also converge

But if I use $b_k=\sin\frac{1}{k}$, $\frac{a_k}{b_k}\to 0$. And here, $b_k$ diverges, $a_k$ must diverge also...

Help?

## 2 Answers

If $\lim_{k\to\infty}\frac{a_k}{b_k}=0$ and $b_k$ converges, then so does $a_k$. If $a_k$ diverges, then so does $b_k$. But in your case the test is inconclusive. So you have no contradiction.

• Ok, that clears things up a lot. I didn't realise that the limit tending to 0 was a special case. Thanks! – user51327 Dec 5 '12 at 16:05

Your series converges by your first example. The second one doesn't work as the limit is zero, and it must be greater than zero and less than infinity.