Is this proof for a sequence convergence problem correct?

[jgens]

Homework Statement

$$\lim_{n\to\infty}a_n=l \rightarrow \lim_{n\to\infty}\frac{a_1+\dots+a_n}{n}=l$$

Homework Equations

N/A

The Attempt at a Solution

Could someone verify that this proof works? I would really appreciate it.

Proof: Since the sequence $\{a_n\}$ converges to $l$, for any given $\varepsilon>0$ it's possible to find a number $N>0$ such that if $n>N$, then $|a_n-l|<\varepsilon/2$. Now, because there are only finitely many numbers $|a_1-l|,\dots,|a_N-l|$, we can choose the greatest such number. Denote this number by $M$.

Suppose that $n>\max{(N,\frac{2MN}{\varepsilon})}$, in which case, it clearly follows that $\frac{\varepsilon}{2}>\frac{MN}{n}$. Therefore,

$$\left| \frac{a_1+\dots+a_N}{n}-\frac{Nl}{n}\right|\leq\frac{|a_1-l|}{n}+\dots+\frac{|a_N-l|}{n}\leq\frac{MN}{n}<\frac{\varepsilon}{2}$$

Moreover, since $n>N$, we also have that

$$\left| \frac{a_{N+1}+\dots+a_n}{n}-\frac{(n-N)l}{n}\right|\leq\frac{|a_{N+1}-l|}{n}+\dots+\frac{|a_n-l|}{n}<\frac{(n-N)\varepsilon}{2n}<\frac{\varepsilon}{2}$$

Combining these two results, we see that

$$\left| \frac{a_1+\dots+a_n}{n}-l\right|\leq\left| \frac{a_1+\dots+a_N}{n}-\frac{Nl}{n}\right|+\left| \frac{a_{N+1}+\dots+a_n}{n}-\frac{(n-N)l}{n}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$

Completing the proof.

 

[jgens]

Bump.

 

[snipez90]

Looks good to me.

 

[jgens]

Thanks!