# Ratio and Root Tests

• August 4th 2010, 07:29 AM
MechEng
Ratio and Root Tests
Good morning All,

I am working on ratio and root tests for infinite series. I would like to post what I have done, and see if anyone agrees with it. Please see below:

$
\displaystyle\sum_{n=2}^{\infty}(-1)^n\frac{123^n}{n!}
$

By the Ratio Test...

$
\displaystyle\lim_{n\rightarrow\infty}\mid\frac{(-1)^{(n+1)}123^{(n+1)}}{(n+1)!}\frac{n!}{(-1)^{n}123^{n}}\mid
$

And...

$
\displaystyle\lim_{n\rightarrow\infty}\frac{123n!} {(n+1)!} = \lim_{n\rightarrow\infty}\frac{123n!}{(n+1)n!} = \lim_{n\rightarrow\infty}\frac{123}{n+1} = 0
$

Therefore...

The orignal series $\displaystyle\sum_{n=2}^{\infty}(-)^n\frac{123^n}{n!}$ is Absolutely Convergent.

Are we good here?

My second problem is:

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

I'm not really sure where to start here. Would I apply the root test to this problem?
• August 4th 2010, 07:33 AM
roninpro
The first problem looks good to me, and you have the right idea for the second. What do you get when you try to take the $n$-th root?
• August 4th 2010, 08:17 AM
MechEng
Thanks!

When I try to take the nth root i get the following:

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

$\displaystyle\lim_{n\rightarrow\infty}|[(-1)^n(\sqrt{n+1}-\sqrt{n})]^{\frac{1}{n}}|$

$=\displaystyle\lim_{n\rightarrow\infty}|(-1)^{\frac{n}{n}}((n+1)^{\frac{1}{2n}}-(n)^{\frac{1}{2n}})|$

$=\displaystyle|(-1)^1((n+1)^0-(n)^0)|$

$=\displaystyle|(-1)((1)-(1))|$

$=\displaystyle|-1+1|$

$=\displaystyle|0|$

$=\displaystyle0$

That can't be right??? I have three problems; one is supposed to be divergent, one conditionally convergent, and one absolutely convergent. Since my first problem was absolutely convergent (and I was able to follow the entire process for the first problem) I would have thought that the second problem would be either conditonally convergent or divergent.

...so, where did I go wrong?
• August 4th 2010, 08:39 AM
Plato
Have you studied the alternating series test? If so that is what is to be used.
• August 4th 2010, 08:56 AM
MechEng
I have reviewed the alternating series test.

I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem.

Is this reasonable?
• August 4th 2010, 09:05 AM
Plato
Quote:

Originally Posted by MechEng
I have reviewed the alternating series test.

I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem.
Is this reasonable?

Did you notice that $\displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?$
• August 4th 2010, 09:18 AM
MechEng
Quote:

Originally Posted by Plato
Did you notice that $\displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?$

Shoot... I did not notice that. Let me try this again...
• August 4th 2010, 09:41 AM
MechEng
When I try to take the nth root i get the following:

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

$=\displaystyle\lim_{n\rightarrow\infty}|[\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}]^{\frac{1}{n}}|$

$=\displaystyle\lim_{n\rightarrow\infty}|\frac{(-1)^\frac{n}{n}}{(n+1)^\frac{1}{2n}+(n)^\frac{1}{2n }}}|$

$=\displaystyle|\frac{(-1)^1}{(n+1)^0+(n)^0}|$

$=\displaystyle|\frac{-1}{1+1}| = \frac{1}{2}$

Am I losing it again somewhere?

Or, are you implying that I use the Ratio Test?

Thanks for the help.
• August 4th 2010, 10:10 AM
Plato
Use the alternating series test.
• August 4th 2010, 11:24 AM
MechEng
Ok, I am waiting to hear back with respect to whether or not using the alternating series test is acceptable.

In the interim, I have worked out the last of three problems and concluded that:

$\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{(2n)!}{3^n(n!)}$

By Ratio Test...

$\displaystyle\lim_{n\rightarrow\infty}|\frac{(-1)^{(n+1)}(2(n+1))!}{3^{(n+1)}((n+1)!)}\frac{3^n(n !)}{(-1)^n(2n)!}|$

$=\displaystyle\lim_{n\rightarrow\infty}\frac{4n^2+ 6n+2}{3n+3} = \infty$

Therefore...

By the Ratio test, the series Diverges

So, this leaves me with my remaining series that is conditionally convergent. And, this means that if I am able to use the Ratio or Root Test, I should wind up with a limit equal to one... right?
• August 4th 2010, 11:31 AM
MechEng
Plato,

I am noticing a trend with the lesson plans that I am following; the questions seem to be quite difficult using the methods covered in that section, but following section will introduce a method that makes the problem quite simple to do. I assume this one is the same.

I am not a huge fan of the way the text book is organized, but i am nearly done with it so it's not worth complaining now.

I appreciate your patience. I used to tutor physics in college, so I know it can be trying when someone keeps mising the fundamental ideas. Thank you for continuing to work with me.
• August 4th 2010, 12:09 PM
MechEng
All tests are now available to use on this problem, so I will heed you advice, Plato, and try the alternating series test...
• August 4th 2010, 12:22 PM
Plato
You should find that it in conditional but not absolute.
To do the later, use the limit comparison test using $\dfrac{1}{\sqrt{n}}$
• August 4th 2010, 05:14 PM
MechEng
Easy question I hope, but...

Since one of the requirements for the limit comparison test is that the series are both positive, does that mean that my original series becomes...

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1 }+\sqrt{n}}$
• August 4th 2010, 05:31 PM
MechEng
Please correct me if I am out of whack here...

Let $\displaystyle\sum a_n = \frac{1}{\sqrt{n+1}+\sqrt{n}}$

So...

$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq rt{n+1}+\sqrt{n}}=0$

By Alternating Series Test... Convergent

Let $\displaystyle\sum b_n = \frac{1}{\sqrt{n}}$ A Divergent P-series

So...

$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq rt{n+1}+\sqrt{n}}\frac{\sqrt{n}}{1}$

$=\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\s qrt{n+1}}=0$

By Limit Comparison Test... Divergent

This doesn't make sense, does it?

Would it make ore sense if the order of the tests was reversed?