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Let $\sum\limits_{n=1}^\infty a_k$ be a convergent series, then the following statement is true:

$$\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^n a_{k} = 0$$

Since $\frac{1}{n}$ is a null series and the partial sum will never get greater than $\sum\limits_{n=1}^\infty a_{n} = c$, the limit is obviously $0$.

How do we argument with this, so that it is sufficient for a proof?

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  • $\begingroup$ The partial sum can definitely get greater than $c$. It might be easier for you to reason about this problem if you define $A_n = \sum_{i=1}^n a_i$ and then consider $(A_n)$ as a sequence, instead of $(a_i)$ as a series. $\endgroup$ – Mees de Vries Dec 6 '17 at 10:06
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The limit of the product of two converging sequences is the product of the limits of the sequences.

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In $\varepsilon$ form, given $$S_n=\sum\limits_{k=1}^{n}a_k$$ $$\lim\limits_{n\rightarrow\infty}S_n=c$$ we have $$\left| S_n - c \right|<\varepsilon$$ from some $n$ onwards. Then $$\left| \frac{S_n}{n} - 0 \right|=\left| \frac{S_n}{n} -\frac{c}{n}+\frac{c}{n}- 0 \right|\leq\left| \frac{S_n}{n} -\frac{c}{n}\right|+\left|\frac{c}{n}- 0 \right|<\frac{\varepsilon}{n}+\left|\frac{c}{n}\right|=\\ \frac{1}{n}\left(\varepsilon+|c|\right)$$ and this goes to $0$ according to squeeze theorem.

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The series converges then $a_n\to 0$ then using this: General Cesaro $\lim\limits_{n\to\infty} \frac{1}{\sum_\limits{k=0}^{n}\lambda_k}\sum_\limits{k=0}^{n}\lambda_k a_k =\lim\limits_{n\to\infty} a_n$

we conclude $$\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^n a_{k} = 0$$

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