1. ## Convergence problem

Show that if $\sum_{n=0}^\infty a_n$ converges conditionally, then $\sum_{n=0}^\infty n^2a_n$ diverges.

Just by looking, I think $n^2a_n$ is unbounded so the second summation diverges, but I don't know how to show this. I know $a_n \leq c$ as a result of convergence so my best attempt so far is to say $a_n/n \leq c/n \leq cn^2$ but I don't know if I'm going in the right direction.

2. Suppose that $\sum n^2a_n$ converges. Then you have that $\lim_{n\to\infty}|n^2a_n|\to{0}$. Now choose $\epsilon=1$, then there exists $N$ such that $n>N\Longrightarrow |n^2a_n|<1\Leftrightarrow |a_n|<\frac{1}{n^2}$. But this contradicts the fact that $\sum a_n$ converges conditionally. Just in case you don't see it, use the comparison test.

Using this approach you can conclude that $\sum_{n=0}^{\infty}n^{1+\alpha}a_n$ diverges for $\alpha>0$.

3. Equivalently, if $\sum a_n$ converges, then $\sum \frac {|a_n|}{n^2}$ converges, which is clear by comparison. All I have done is taken contrapositives and renamed the terms.

4. Originally Posted by cubrikal

Show that if $\sum_{n=0}^\infty a_n$ converges conditionally, then $\sum_{n=0}^\infty n^2a_n$ diverges.

Just by looking, I think $n^2a_n$ is unbounded so the second summation diverges, but I don't know how to show this. I know $a_n \leq c$ as a result of convergence so my best attempt so far is to say $a_n/n \leq c/n \leq cn^2$ but I don't know if I'm going in the right direction.
suppose $\sum n^2a_n$ is convergent. then $\lim_{n\to\infty} n^2a_n=0$ and hence $\lim_{n\to\infty} \frac{|a_n|}{\frac{1}{n^2}}=0.$ but then, by the limit comparison test, $\sum |a_n|$ will be convergent. contradiction!