# Proove that convergence of one series implies the convergence of another series

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• Dec 1st 2009, 02:05 PM
Pythagorean12
Proove that convergence of one series implies the convergence of another series
• Dec 1st 2009, 02:57 PM
Plato
Quote:

Originally Posted by Pythagorean12

Pythagorean12, this is the third or fourth question you have posted.
But in each case you have shown absolutely no effort on your part.
That is not the way this forum works.
This is not a homework service, they charge.
This is a free homework help service.
We help once you show some effort.
• Dec 1st 2009, 03:02 PM
Pythagorean12
• Dec 1st 2009, 03:06 PM
Plato
Quote:

Originally Posted by Pythagorean12

That is nonsense. Take the example $a_n=\frac{1}{n}$.
• Dec 1st 2009, 03:23 PM
Focus
Pick an epsilon >0, then we have N such that $n>N \implies |a_n-a|<\epsilon$.

Now $|(a_1+...a_n)/n-a|=\frac{1}{n}|a_1+...+a_n-na|\leq|\frac{1}{n}|a_1-a|+...+\frac{1}{n}|a_n-a|$.

You are home and dry if $\frac{1}{n}(|a_1-a|+...+|a_N-a|)<\epsilon$ as the other terms are less than $\frac{\epsilon}{n}$.

So lets increase N until we can minimise the first terms, note that for n>N we have 1/n<1/N, so we wish to find K such that for n>K
$\frac{1}{K}(|a_1-a|+...+|a_N-a|)<\epsilon$ and a wise choice is $K=[\frac{2}{\epsilon}(|a_1-a|+...+|a_N-a|)]$ where [] denotes the next integer up.

Let $M=max\{N,K\}$ then for n>M we have that
$
|b_n-a|\leq|\frac{1}{n}(|a_1-a|+...+|a_N-a|)+\frac{1}{n}|a_{N+1}-a|+...+\frac{1}{n}|a_n-a|<\epsilon
$
or some constant multiple.