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 $c=0$.


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.

  • $\begingroup$ 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

I believe that this is true with each $a_1 = 1$. We note:

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.