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

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

Let and let . Show that, if as , then also.

2. Originally Posted by Pythagorean12
Let and let . Show that, if as , then also.
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.

3. As , Assuming Making, . Thus .

I am sorry if it looks like the solution of an absolute idiot - I am absolutely new to this part of mathematics.

4. Originally Posted by Pythagorean12
As , Assuming Making, . Thus .

I am sorry if it looks like the solution of an absolute idiot - I am absolutely new to this part of mathematics.
That is nonsense. Take the example $\displaystyle a_n=\frac{1}{n}$.

5. Pick an epsilon >0, then we have N such that $\displaystyle n>N \implies |a_n-a|<\epsilon$.

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

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

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
$\displaystyle \frac{1}{K}(|a_1-a|+...+|a_N-a|)<\epsilon$ and a wise choice is $\displaystyle K=[\frac{2}{\epsilon}(|a_1-a|+...+|a_N-a|)]$ where [] denotes the next integer up.

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