7

As in the title, consider the problem:

If $\sum n a_n$ converges, then $\sum a_n$ converges and $\sum |a_n|^p$ converges for $p>1$.

My intuition for the first part of the proof is to use this equality,

$$\sum a_n = \sum n a_n -\big(\sum a_n(n-1)\big) $$

Obviously the first part of the RHS converges by assumption, and I want to say the second part of the RHS does as well, however I don't know if it's entirely obvious that it does. Is this reasoning valid? And if there is a more clever way, please let me know.

Additionally, I'm not sure where to begin in showing that the hypothesis implies $\sum|a_n|^p$ converges.

All help is greatly appreciated.

sequenceDerivative
  • 353

3 Answers3

6

The reasoning presented is not valid; the two expressions are equal only if you can rearrange terms, which requires $\sum a_n$ to be absolutely convergent (which it need not be).

Part 1: Use Abel's test, with $b_n=\frac{1}{n}$.

Part 2: You might begin by noting that since $\sum na_n$ converges, we must have $\lim_{n\to\infty} na_n=0$.

vadim123
  • 82,796
5

For part 2: Since $\lim_{n \rightarrow \infty} na_n = 0$, there must exist some $N$ such that for $n \geq N$, $|na_n| \leq 1$. This implies $|a_n| \leq \frac{1}{n}$, and therefore $|a_n|^p \leq \frac{1}{n^p}$, and you are done by the comparison test.

Alex Zorn
  • 4,304
5

  • Use the summation by parts $$ \sum_{n=1}^N a_n = \sum_{n=1}^N na_n \frac{1}{n} = \frac{1}{N}\sum_{n=1}^N n a_n + \sum_{n=1}^{N-1} (\frac{1}{n}-\frac{1}{n+1})\sum_{m=1}^n m a_m$$ note that $|\sum_{m=1}^n m a_m| < B$ and $\frac{1}{n}-\frac{1}{n+1} = \frac{1}{n (n+1)}$

  • For the second one it is trivial : $\sum_{n=1}^\infty n a_n$ converges means that $n a_n \to 0$ so that $n^p |a_n|^p \to 0$ i.e. $|a_n|^p < C n^{-p}$ and $\sum_{n=1}^\infty |a_n|^p $ converges for $p > 1$

reuns
  • 77,999