Suppose $\sum{a_n}$ is a convergent series of real numbers. Either prove that $\sum{b_n}$ converges or give a counter-example, when we define $b_n$ by:

  1. $a_n \sin(n)$
  2. $n^{\frac{1}{n}}a_n$

For the first one, I was thinking of using the fact that $|\sin(n)| \leq 1$ and then using comparison test. However, we don't know that $\sum{|a_n|}$ converges.

For the second one, I was thinking of using the fact that $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$. But, I'm completely stuck.



For 1), take $a_n = \dfrac{\sin n}{n}$. Then $\sum a_n$ converges (by Dirichlet's test), but $$ \sum \frac{\sin^2 n}{n} $$ diverges, see Convergence of $\sum_{n=1}^\infty \frac{\sin^2(n)}{n}$.

For 2), the series $\sum b_n$ converges by Abel's test

  • $\begingroup$ Could you please just sketch how to use Abel's test? I don't see how it applies here. $\endgroup$ – Alex M. Nov 11 '15 at 21:50
  • $\begingroup$ @AlexM. $(n^{1/n})$ is monotone (for $n \ge 3$) and bounded. (Maybe you're thinking of the "other" Abel's test for power series. The terminology seems to be a little non-standardized.) $\endgroup$ – mrf Nov 11 '15 at 21:52
  • $\begingroup$ I hadn't checked your link and I had a different statement in mind. $\endgroup$ – Alex M. Nov 11 '15 at 21:53
  • $\begingroup$ @AlexM. Thank you so much! I wanted to use Abel's test on the second one, but I didn't realize you can throw out the first n terms that make the sequence not monotone as long as after those n terms the sequence is monotone (is that the correct way of thinking)? $\endgroup$ – jwise Nov 12 '15 at 0:32
  • $\begingroup$ @jwise: Yes, that is correct, but I think that you wanted to talk to Mrf, not to me. $\endgroup$ – Alex M. Nov 12 '15 at 7:07

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