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

**Pythagorean12** · December 1st 2009, 02:05 PM

Let

and let

. Show that, if

as

, then

also.

**Plato** · December 1st 2009, 02:57 PM

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.

**Pythagorean12** · December 1st 2009, 03:02 PM

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.

**Plato** · December 1st 2009, 03:06 PM

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 $a_n=\frac{1}{n}$ .

**Focus** · December 1st 2009, 03:23 PM

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
$<br /> |b_n-a|\leq|\frac{1}{n}(|a_1-a|+...+|a_N-a|)+\frac{1}{n}|a_{N+1}-a|+...+\frac{1}{n}|a_n-a|<\epsilon<br />$ or some constant multiple.

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