# Series: Absolutely convergent, con convergent or divergent

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• February 21st 2009, 08:54 AM
thehollow89
Series: Absolutely convergent, con convergent or divergent
The series from 1 to infinity of n!/n^n? I tried the ratio and root test and they didn't work, unless I messed up. I was thinking a comparison test, but I'm not sure if you can verify something is absolutely convergent using the comparison test or limit comparison test.
• February 21st 2009, 08:59 AM
Moo
Hello,
Quote:

Originally Posted by thehollow89
The series from 1 to infinity of n!/n^n? I tried the ratio and root test and they didn't work, unless I messed up. I was thinking a comparison test, but I'm not sure if you can verify something is absolutely convergent using the comparison test or limit comparison test.

The term in the series is positive, so here, absolute convergence and convergence are equivalent.
Use the ratio test :
$a_n=\frac{n!}{n^n}$

$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!}=\frac{(n+1) n^n}{(n+1)^n (n+1)}=\frac{n^n}{(n+1)^n}$
Now you have to take the limit as n goes to infinity :
$\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n=\ left(\frac{n+1}{n}\right)^{-n}=\left(1+\frac 1n\right)^{-n}$

now use the fact that $\lim_{n \to \infty} \left(1+\frac 1n\right)^n=e>1$
• February 21st 2009, 09:04 AM
thehollow89
So it also goes to e if the nth power is going to negative infinity? I was thinking something similar, but my notes are probably wrong I'm now thinking.
• February 21st 2009, 09:05 AM
Moo
Quote:

Originally Posted by thehollow89
So it also goes to e if the nth power is going to negative infinity? I was thinking something similar, but my notes are probably wrong I'm now thinking.

$\left(1+\frac 1n\right)^{-n}$ is just another way for writing $\frac{1}{\left(1+\frac 1n\right)^n}$ (Wink)
• February 21st 2009, 09:07 AM
thehollow89
Wouldn't that make it 1/e then?
• February 21st 2009, 09:09 AM
Moo
Quote:

Originally Posted by thehollow89
Wouldn't that make it 1/e then?

Yes it does... I never said it didn't (Surprised)
But with knowing these limits, you can say whether the ratio test fails or not.
• February 21st 2009, 09:11 AM
thehollow89
Quote:

Originally Posted by Moo
Yes it does... I never said it didn't (Surprised)
But with knowing these limits, you can say whether the ratio test fails or not.

Okay just making sure. Thank you for your help, I hate making dumb mistakes. It's normal in Math to learn from mistakes right? lol
• February 21st 2009, 09:13 AM
Moo
Quote:

Originally Posted by thehollow89
Okay just making sure. Thank you for your help, I hate making dumb mistakes. It's normal in Math to learn from mistakes right? lol

Yes, of course ! But there is no mistake if you don't try ^^
Next time, it would be better if you post what you would have done so far (Wink)
• February 21st 2009, 09:16 AM
thehollow89
Quote:

Originally Posted by Moo
Yes, of course ! But there is no mistake if you don't try ^^
Next time, it would be better if you post what you would have done so far (Wink)

I will next time, I think it'd be better if I learn some LaTeX? I think that's what people use to post the Math symbols and what not.