(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]\lim_{n\to\infty}a_n=l \rightarrow \lim_{n\to\infty}\frac{a_1+\dots+a_n}{n}=l[/tex]

2. Relevant equations

N/A

3. The attempt at a solution

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

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

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

[tex]\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}[/tex]

Moreover, since [itex]n>N[/itex], we also have that

[tex]\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}[/tex]

Combining these two results, we see that

[tex]\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[/tex]

Completing the proof.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Sequences proof problem

**Physics Forums | Science Articles, Homework Help, Discussion**