Question

Prove that if \[z _{n}\rightarrow A \] as\[x \rightarrow \infty\] then \[\lim_{n \rightarrow \infty}(z _{1}+...+z _{n})/n = A\]

Answer 1

I have an idea , but not sure if it headed in a good direction.... Here it goes anyways
The sequens can be writen like this:\[z _{1}/n+z _{2}/n+...+z _{n}/n\]
as n gets bigger its terms are getting as close as you need to this series:
\[nA/n\]
which has a limit A

Answer 2

sry, where i wrote "series", i ment sequence

Answer 3

use the definition


since \[z_n\to A\] there exists an \(N\) such that for all \(n>N\) we have

\[|A-z_n|<\epsilon\]

break \[z_1+z_2+
\cdots+z_n\] into

\[(z_1+z_2+
\cdots z_N)+(z_{N+1}+\cdots+z_n)\]


and use the triangel inequality

Answer 4

@Zarkon

Answer 5

sry, just to make sure i understand it right:
so my aprouch is right. Just I have to put it in epsilon delta (n>N) way. First N terms will vanish anyways if N is big enough. Correct me if i am wrong