# Limit of a sequence implication [duplicate]

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. FalconDec 29 '16 at 13:38

• Please give us some insight into where you are stuck, what you have tried, etc.. – Tom Nov 10 '13 at 18:41
• Do you know Cesaro's-Stolz lemma? Check this en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem – passenger Nov 10 '13 at 18:42
• you can prove this by the definition of the limit directly. If you can not write the whole proof, you can tell me, and then I can post it to you. – Lei Li Nov 10 '13 at 18:44
• No, but Cesàro mean @ passenger – surry Nov 10 '13 at 18:47
• – Martin Sleziak Dec 29 '16 at 12:27

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 $n\to \infty$