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**SlipEternal** The coefficients follow $\displaystyle \dfrac{2n+1}{n(n+1)}$ when $\displaystyle n$ is even. But not when it is odd. Anyway, if you can show that the coefficients are no greater than 1, then you can show that $\displaystyle 0\le \sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k=n}^{2n-1} a_k \right) = \sum_{n\ge 1} b_na_n$ where $\displaystyle b_n = \begin{cases}\dfrac{2n+1}{n(n+1)} & n\mbox{ is even} \\ ??? & n\mbox{ is odd}\end{cases} \le 1$. Hence, $\displaystyle \sum_{n\ge 1} b_na_n \le \sum_{n\ge 1} a_n$