1. Check series convergence 2

I have SUM [ ((-1)^n * n!)/n^n ] from n=1 to +inf . I need to check both absolute and conditional convergence.

Now this is very tricky.. the (-1)^n tells me i need to try with the alternating series test.. Than An= n!/n^n , and lim n->inf An is impossible to find for me.

When i tried with Ratio test for absolute convergence i got 1 as result which tells me nothing .

My question here is: Can i cancel n! , and say the series will converge is the series SUM [ 1/n^n ] from n=1 to +inf is converging too ? The last series can be considered as p-series or a geometric series , and it converges since n will always be positive .

At the end does that tells me that the first series is also absolute convergent .

2. Re: Check series convergence 2

Originally Posted by MirceM
I have SUM [ ((-1)^n * n!)/n^n ] from n=1 to +inf . I need to check both absolute and conditional convergence.
$\displaystyle \frac{{(n + 1)!}}{{{{(n + 1)}^{n + 1}}}} \cdot \frac{{{n^n}}}{{n!}} = {\left( {1 - \frac{1}{{n + 1}}} \right)^n} \to {e^{ - 1}}$

Tells us all we need to know.

3. Re: Check series convergence 2

@Plato

Thats the result i got too (not including e^-1) , because when n->+inf the result is 1 . I cant understand where is e^-1 from..

4. Re: Check series convergence 2

Originally Posted by MirceM
^ Thats the result i got too (not including e^-1) , because when n->+inf the result is 1 . I cant understand where is e^-1 from..
Suppose that $\displaystyle b\ne 0~\&~c\in\mathbb{Z}^+$ then $\displaystyle {\left( {1 + \frac{b}{{n + c}}} \right)^n} \to {e^b}$.

5. Re: Check series convergence 2

Yes yes .. i got it now... i hope this is the only "special case" of limits when n->inf that is a bit confusing for noobs in math like me.. Because in any other normal example i`m canceling some of the conditions and than just replace n as inf and calculate the result, like i did in this one. I mean if you dont know that term from past , you will go just str8 forward..