Test Series For Convergence

**boquist** · August 15th 2010, 12:37 PM

I cannot figure out how to write this series in formula summation form. I know it is an alternating series. So far I have:
a_n(n=0 to infiniti) = (-1)^n(3^2n)/(something)

I cannot determine the denominator...

I believe there was also a mistake on the printout of the problem. The third term of the series is supposed to be (2^2)(4^2).

Thank you for any help!

**Vlasev** · August 15th 2010, 01:56 PM

Here is my way of doing it.For now ignore the first term, that is for n=0. The second term is for n=1, etc

So you have for n=3 you have

$\displaystyle \frac{3^{2.3}}{2^24^26^2} = \frac{3^{2.3}}{2^24^2(2.3)^2)}$

So the general terms should be:

$\displaystyle\frac{3^{2n}}{2^24^2\ldots (2n)^{2}} = \frac{3^{2n}}{2^2.1^2.2^2.2^.2^2.3^2\ldots 2^2n^2} = \frac{3^{2n}}{2^2.2^2.2^2\ldots 1^22^23^2\ldots n^2} = \frac{3^{2n}}{2^{2n}(n!)^2}$

Oh, and of course there is the negative sign! Now if you plug in k = 0 in the above, you get 1 so we have taken care of the first term as well!

Now you need to test the series for convergence:

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n(3/2)^{2n}}{(n!)^2}$

There are many ways you can test this for convergence and it should be pretty easy. I'm thinking comparison test with something with just a factorial on the bottom. Or you can compare successive terms.

**boquist** · August 15th 2010, 03:09 PM

Thanks for the reply and the help.
I used the Alternating Series Test and the Squeeze Theorem.
This is what I came up with, maybe you can tell me if you agree with my answer?

Thanks again!

**Vlasev** · August 15th 2010, 03:21 PM

No you cannot do that! From what you have written I get that you only show that the term of the series goes to 0 as n approaches infinity. What you really need to show is that the partial sums converge to some finite number. What you have written will help to show that though.

Here is how I would do it:

[LaTeX ERROR: Convert failed]

So we have the series [LaTeX ERROR: Convert failed] has its every term greater than the series you have. You may recognize this series as the series for the exponential $e^{x}$ with $x = 9/4$ . So the value of this series is $e^{9/4}$ which is definitely a finite value. Hence by the series comparison test we have that your series converges.

**yeKciM** · August 15th 2010, 03:24 PM

for series type : $\displaystyle \sum _{n=0} ^{\infty} (-1)^n a_n$

u can use :

1° $\displaystyle \lim _{n\to\infty} {a_n} =0$
2° $a_n>a_{n+1}$

if those are true, then series converge

sorry about this post here... please can someone delete it

didn't see that he is doing what i wrote here .... sorry

**Vlasev** · August 15th 2010, 03:26 PM

Oh, I see what you've done there. I missed that he is doing the alternating series test.

**chisigma** · August 15th 2010, 10:21 PM

Originally Posted by Vlasev

Now you need to test the series for convergence:

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n(3/2)^{2n}}{(n!)^2}$

There are many ways you can test this for convergence and it should be pretty easy. I'm thinking comparison test with something with just a factorial on the bottom. Or you can compare successive terms.

The series...

$\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{(\frac{3}{2})^{2n}}{(n!)^{2}}$ (1)

... is of the 'alternating sign' type and it converges if the general term in modulus tends monotonically to 0 if n tends to infinity. Now if You remember the Stirling approximation...

$\displaystyle n! \sim \sqrt{2 \pi} \ n^{n+\frac{1}{2}}\ e^{-n}$ (2)

... You find that...

$\displaystyle (n!)^{2} \sim 2 \pi \ n^{2n+1}\ e^{-2n}$ (3)

... so that is...

$\displaystyle \frac{(\frac{3}{2})^{n}}{(n!)^{2}} \sim \frac{(\frac{3}{2})^{n}\ e^{2n}}{2 \pi\ n^{2n+1}}$ (4)

The general term in modulus tends monotonically to 0 if n tends to infinity: the series converges...

Kind regards

$\chi$ $\sigma$

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