# testing for convergence/divergence

• Mar 22nd 2010, 11:56 PM
Dr Zoidburg
testing for convergence/divergence
I've got a few problems here I'm wanting some guidance over.
1. $\displaystyle \sum_{n=0}^{\infty} \frac{n!}{2^{n^2}}$
Am I right in thinking the Ratio test is the one to use here?
I get
$\displaystyle \frac {n+1}{2^{2n+1}}$
Do I use the Ratio Test again? I end up with $\displaystyle \frac {1}{2}$

2. $\displaystyle \sum_{n=1}^{\infty} \frac{2^n +1}{3^n -cosn}$
Absolute convergence test?
|cos n|<1, thus the above series can be reduced down to $\displaystyle (\frac {2}{3})^n = \frac{2}{3}$
>>edit: sorry, made a mistake in the original post in the series equation. Now rectified.<<

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

I'm thinking the Root test here?
then, since $\displaystyle \lim_{n \to\infty} {n}^\frac{1}{n} = 1$, we're left with just $\displaystyle (-1)^n$

Any/all of these on the right track or have I totally lost the plot here?
Cheers!
• Mar 23rd 2010, 01:44 AM
Prove It
Quote:

Originally Posted by Dr Zoidburg
I've got a few problems here I'm wanting some guidance over.
1. $\displaystyle \sum_{n=0}^{\infty} \frac{n!}{2^{n^2}}$
Am I right in thinking the Ratio test is the one to use here?
I get
$\displaystyle \frac {n+1}{2^{2n+1}}$
Do I use the Ratio Test again? I end up with $\displaystyle \frac {1}{2}$

2. $\displaystyle \sum_{n=1}^{\infty} \frac{2n+1}{3-cosn}$
Absolute convergence test?
|cos n|<1, thus the above series can be reduced down to $\displaystyle (\frac {2}{3})^n = \frac{2}{3}$

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

I'm thinking the Root test here?
then, since $\displaystyle \lim_{n \to\infty} {n}^\frac{1}{n} = 1$, we're left with just $\displaystyle (-1)^n$

Any/all of these on the right track or have I totally lost the plot here?
Cheers!

1. $\displaystyle \sum_{n=0}^{\infty} \frac{n!}{2^{n^2}}$

Yes, use the ratio test.

You need to find

$\displaystyle \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{\frac{(n + 1)!}{2^{(n + 1)^2}}}{\frac{n!}{2^{n^2}}}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{(n + 1)!2^{n^2}}{n!2^{(n + 1)^2}}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{(n + 1)2^{n^2}}{2^{n^2 + 2n + 1}}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{n + 1}{2^{2n + 1}}\right|$

$\displaystyle = \lim_{n \to \infty}\frac{n + 1}{2^{2n + 1}}$ since it's a positive term series

$\displaystyle = \lim_{n \to \infty}\frac{1}{2\ln{2}\cdot 2^{2n + 1}}$ by L'Hospital's Rule

$\displaystyle = 0$.

Since this limit $\displaystyle < 0$, the series is convergent.
• Mar 23rd 2010, 03:36 AM
Dr Zoidburg
thanks for that.
Out of curiosity, can I just do a Ratio test again instead of use L'Hopital's Rule?
• Mar 23rd 2010, 04:44 AM
Prove It
Quote:

Originally Posted by Dr Zoidburg
thanks for that.
Out of curiosity, can I just do a Ratio test again instead of use L'Hopital's Rule?

No, because you don't get another series. You get a term that you're finding the limit of...