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:
- $a_n \sin(n)$
- $n^{\frac{1}{n}}a_n$
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.
Thanks!
Answer
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
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