# Thread: Convergence of a Series

1. ## Convergence of a Series

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.

2. Hi

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

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

3. Originally Posted by flyingsquirrel
Hi

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

This series is alternating hence you can try to show the convergence using the alternating series test.
Be careful with that n not k...

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

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

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

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

4. Originally Posted by Mathstud28
Be careful with that n not k...

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

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

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

and also since $\displaystyle \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{\sq rt{n+5}}=0$ 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

5. 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

6. 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

7. 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 $\displaystyle a_{n+1}<a_n,\forall{n}>0$ Isnt enough?

8. Originally Posted by Mathstud28
So you are saying that $\displaystyle a_{n+1}<a_n,\forall{n}>0$ Isnt enough?
That is not what you had in your post you had that the $\displaystyle \lim_{n \to \infty}|a_n|=0$

But neither of the conditions:

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

or

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

on its own is enough you need both.

RonL

9. Originally Posted by Mathstud28
Be careful with that n not k...

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

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

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

and also since $\displaystyle \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{\sq rt{n+5}}=0$ 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 $\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n+5}}=\lim_{n\to\i nfty}\bigg|\frac{(-1)^{n}}{\sqrt{n+5}}\bigg|$

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

10. 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 $\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n+5}}=\lim_{n\to\i nfty}\bigg|\frac{(-1)^{n}}{\sqrt{n+5}}\bigg|$

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

RonL