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The limit of the sequence is zero

Let \{a_n\}_{n \in \mathbb{N}} be a real valued sequence such that the series \displaystyle \sum \limits_{n=1}^{\infty} \frac{a_n}{n^s} converges. Prove that

\displaystyle \lim_{n \rightarrow +\infty} \frac{a_1+a_2+\cdots+a_n}{n^s} =0

Solution

For every n \in \mathbb{N} let us set \displaystyle R_n=\sum_{k=n}^{\infty} \frac{a_k}{k^s} thus a_k = k^s \left ( R_k - R_{k+1} \right ) . Since the series converges we conclude that R_n \rightarrow 0 and since S_n =\sup \limits_{k \geq n} \left | R_k \right | is finite forall  n \in \mathbb{N} we deduce that S_n \rightarrow 0 . Thus:

\begin{aligned} \left\lvert \frac{a_1 + \cdots + a_n}{n^s}\right\rvert &= n^{-s}\bigg\lvert \sum_{k=1}^n k^s(R_k - R_{k+1})\bigg\rvert\\ &= n^{-s}\bigg\lvert \sum_{k=1}^n k^s R_k - \sum_{k=2}^{n+1} (k-1)^s R_k \bigg\rvert\\ &= n^{-s}\left\lvert R_1 - n^s R_{n+1} + \sum_{k=2}^n \left(k^s - (k-1)^s\right)R_k\right\rvert\\ &\leq \frac{S_1}{n^s} + S_{n+1} + \sum_{k=2}^n \frac{k^s - (k-1)^s}{n^s}S_k\\ &\leq \frac{S_1}{n^s} + S_{n+1} + S_2\sum_{k=2}^{\lfloor\sqrt{n}\rfloor} \frac{k^s - (k-1)^s}{n^s} + S_{\lfloor\sqrt{n}\rfloor+1}\sum_{k=\lfloor\sqrt{n}\rfloor+1}^n\frac{k^s-(k-1)^s}{n^s}\\ &= \frac{S_1}{n^s} + S_{n+1} + S_2\frac{\lfloor\sqrt{n}\rfloor^s-1}{n^s} + S_{\lfloor\sqrt{n}\rfloor+1} \frac{n^s - \lfloor\sqrt{n}\rfloor^s}{n^s}\\ &\leq\frac{S_1}{n^s} + S_{n+1} + \frac{S_2}{n^{s/2}} + S_{\lfloor\sqrt{n}\rfloor+1} \end{aligned}

Hence for a given \epsilon>0 one can pick a n_\epsilon such that

\displaystyle S_{\left \lfloor \sqrt{n_\epsilon } \right \rfloor}<\frac{\epsilon }{4} \quad \text{and} \quad \frac{S_1}{n^{s/2}_\epsilon}<\frac{\epsilon}{4}

Hence \displaystyle{\left | \frac{a_1+a_2+\cdots+a_n}{n^s} \right |<\epsilon} and the result follows.

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