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**SlipEternal** A subsequence is very different from a partial sum. A sum of partial sums may not converge. But, I think I figured it out.

$\displaystyle \sum_{n\ge 1} \dfrac{1}{n}\left( \sum_{k = n}^{2n-1}a_k \right)$

Now, can you show that $\displaystyle \sum_{k = n}^{2n-1} a_k \le na_n$ for almost all $\displaystyle n$?

Since $\displaystyle \sum_{n\ge 1} a_n$ converges and $\displaystyle 0\le a_n$ for all n, it must be that $\displaystyle \lim_{n\to \infty} a_n = 0$. So what I am getting at, $\displaystyle \underbrace{a_n+ a_{n+1} + \ldots + a_{2n-1}}_{n\mbox{ terms}} \le \underbrace{a_n + a_n + \ldots + a_n}_{n\mbox{ terms}}$