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Consider a series $a_n \in {\mathbb R}_+$ ($n \in {\mathbb Z}_+$) with \begin{equation} \lim_{n\rightarrow \infty} a_n = 0. \end{equation}

It is easy to find an example that \begin{equation} \lim_{n\rightarrow \infty} \sum_{i=1}^{n}a_i = \infty. \end{equation} For example $a_n = 1/n$ or $a_n = 1/\log(n)$.

However, does such a $a_n$ exist that $\lim_{n\rightarrow \infty} a_n = 0$ and

\begin{equation} \lim_{n\rightarrow \infty} \frac {1}{n}\sum_{i=1}^{n}a_i = c, \end{equation} where $c > 0$?

PS: For $a_n = 1/n$ or $a_n = 1/\log(n)$, it can be verified that $c=0$.

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If a sequence $(a_n)$ is convergence and $a_n\to L$, it can be shown that $$ \lim_{n\rightarrow \infty} \frac {1}{n}\sum_{i=1}^{n}a_i = L $$ also. So basically, you cannot find a sequence $(a_n)$ such that $lim_{n\to\infty} a_n=0$ while
$\lim_{n\rightarrow \infty} \frac {1}{n}\sum_{i=1}^{n}a_i > 0$

For more information, see Cesàro summation from wikipedia.

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  • $\begingroup$ I've edited my answer to (hopefully) match your question :) $\endgroup$ – BigbearZzz Oct 7 '15 at 3:29
  • $\begingroup$ I'm glad I could help. $\endgroup$ – BigbearZzz Oct 7 '15 at 3:35
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I believe that this is true with each $a_1 = 1$. We note:

$$ \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^n 1 =\lim_{n\rightarrow \infty} \frac{1}{n} n = 1$$

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  • $\begingroup$ Thanks, but $\lim_{n\rightarrow \infty} a_n$ seems not $0$ $\endgroup$ – Ryan Oct 7 '15 at 3:24
  • $\begingroup$ sorry, my question initially confusing, and I have re-edited it $\endgroup$ – Ryan Oct 7 '15 at 3:26
  • $\begingroup$ Sorry, I didn't know you were looking for that condition. Bigbear's answer implies that no such sequence exists. $\endgroup$ – Frank Wang Oct 7 '15 at 3:27

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