Convergence of a Series

**smiler** · May 14th 2008, 08:34 AM

Can anyone help me with the following problem please:

Using convergence tests, determine whether the series

∑ (-1)n/√(n+5) over n=1 and n=∞

converges or diverges.

Many thanks to anyone who can provide me with any advice on solving this problem.

**flyingsquirrel** · May 14th 2008, 08:51 AM

Hi

$\sum_{n=1}^{\infty}(-1)^n\frac{1}{\sqrt{n+5}}$

This series is alternating hence you can try to show the convergence using the alternating series test.

**Mathstud28** · May 14th 2008, 11:52 AM

Originally Posted by flyingsquirrel

Hi

$\sum_{k=1}^{\infty}(-1)^n\frac{1}{\sqrt{n+5}}$

This series is alternating hence you can try to show the convergence using the alternating series test.

Be careful with that n not k...

$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{\sqrt{n+5}}$

Expanding upon Flying squirrels explanation...since $a_{n+1}<a_n$

Where $a_n=\frac{1}{\sqrt{n+5}}$

and also since $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{\sq rt{n+5}}=0$ we can see by the alternating series test that this series is at least conditionally convergent

**CaptainBlack** · May 14th 2008, 12:49 PM

Originally Posted by Mathstud28

Be careful with that n not k...

$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{\sqrt{n+5}}$

Expanding upon Flying squirrels explanation...since $a_{n+1}<a_n$

Where $a_n=\frac{1}{\sqrt{n+5}}$

and also since $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{\sq rt{n+5}}=0$ we can see by the alternating series test that this series is at least conditionally convergent

Alternating series test: If the limit of the absolute value of the terms is 0 and beyond some point the sequence of absolute values is non-increasing [that is there exists an N such that for all n>N; a_n>=a_(n+1)] then the series converges (conditionally).

RonL

**Mathstud28** · May 14th 2008, 02:27 PM

Originally Posted by CaptainBlack

Alternating series test: If the limit of the absolute value of the terms is 0 and beyond some point the sequence of absolute values is non-increasing [that is there exists an N such that for all n>N; a_n>=a_(n+1)] then the series converges (conditionally).

RonL

How is this different than what I said? I didnt state some of the things but what I said was correct

**CaptainBlack** · May 14th 2008, 08:16 PM

Originally Posted by Mathstud28

How is this different than what I said? I didnt state some of the things but what I said was correct

You have not shown convergence untill you have shown that the series is non-increasing beyound some point, and you have not stated that this is needed. Nor have you done the alternative, which is to just say that this is needed but obvious.

RonL

**Mathstud28** · May 14th 2008, 08:28 PM

Originally Posted by CaptainBlack

You have not shown convergence untill you have shown that the series is non-increasing beyound some point, and you have not stated that this is needed. Nor have you done the alternative, which is to just say that this is needed but obvious.

RonL

So you are saying that $a_{n+1}<a_n,\forall{n}>0$ Isnt enough?

**CaptainBlack** · May 14th 2008, 08:55 PM

Originally Posted by Mathstud28

So you are saying that $a_{n+1}<a_n,\forall{n}>0$ Isnt enough?

That is not what you had in your post you had that the $\lim_{n \to \infty}|a_n|=0$

But neither of the conditions:

$\lim_{n \to \infty}|a_n|=0$

or

$\exists N;\ \forall n>N,\ |a_{n+1}| \le |a_n|$

on its own is enough you need both.

RonL

**Mathstud28** · May 15th 2008, 02:59 AM

Originally Posted by Mathstud28

Be careful with that n not k...

$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{\sqrt{n+5}}$

Expanding upon Flying squirrels explanation...since $a_{n+1}<a_n$

Where $a_n=\frac{1}{\sqrt{n+5}}$

and also since $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{\sq rt{n+5}}=0$ we can see by the alternating series test that this series is at least conditionally convergent

if you look closely I had both....I didnt state that n>0 but I had that condition

and I took $\lim_{n\to\infty}\frac{1}{\sqrt{n+5}}=\lim_{n\to\i nfty}\bigg|\frac{(-1)^{n}}{\sqrt{n+5}}\bigg|$

are you saying that my prolem is that I just did it and did not explicity state what I was doing?

**CaptainBlack** · May 15th 2008, 03:35 AM

Originally Posted by Mathstud28

if you look closely I had both....I didnt state that n>0 but I had that condition

and I took $\lim_{n\to\infty}\frac{1}{\sqrt{n+5}}=\lim_{n\to\i nfty}\bigg|\frac{(-1)^{n}}{\sqrt{n+5}}\bigg|$

are you saying that my prolem is that I just did it and did not explicity state what I was doing?

OK

RonL

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