This question already has an answer here:

Be $(a_n)_{n\in N}$ a sequence in $ R $ and $a\in R$. Show that $\lim_{n\to \infty}a_n=a ~~~~~\Rightarrow ~~~~~\lim_{n\to \infty} \left(\frac1n \sum_{k=1}^n a_k\right) = a$


marked as duplicate by Martin Sleziak, Rohan, Behrouz Maleki, Henrik, C. Falcon Dec 29 '16 at 13:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Because $a_n\to a$ there is a $M>0:|a_n|\leq M$ for every $n\geq 1$. Just to make it more simple suppose that $a=0$(because if we set $b_n=a_n-a$ we have that $b_n\to 0$ and thus it's enough to show that $\frac {b_1+b_2+...+b_n}{n}\to 0 $ if $b_n\to 0$).

Let $ε>0$.Because $\frac {M}{\sqrt n}\to 0$ there is a $n_o\in \Bbb N:|a_n|<ε/2$ and $\frac {M}{\sqrt n}<ε/2$ for $n\geq n_0$. So for $n\geq n_0^2$ we have $|\frac {a_1+a_2+...+a_n}{n}|\leq \frac {a_1+a_2+...+a_{\sqrt n}}{n}|+|\frac {a_{\sqrt {n+1}}+...+a_n}{n}|\leq \frac {M[\sqrt n]}{n}+\frac {nε}{2}\frac {1}{n}\leq \frac {M}{\sqrt n}+\frac {ε}{2}\leq \frac {ε}{2}+\frac {ε}{2}=ε$ and thus $\frac {a_1+a_2+...+a_n}{n}\to a$ for $n\to \infty$

  • $\begingroup$ @surry ,check my proof please.thank you $\endgroup$ – Haha Nov 24 '13 at 12:53

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