Do these converge absolutely, conditionally, or diverge?

sigma from n = 1 to infinity of (n + 1)^2 / n!

sigma from n = 1 to infinity of (1/ (arctan(n))^(2n)

explain please!

Printable View

- Nov 28th 2010, 06:35 PMPlaythiousNeed help with these convergence Tests?
Do these converge absolutely, conditionally, or diverge?

sigma from n = 1 to infinity of (n + 1)^2 / n!

sigma from n = 1 to infinity of (1/ (arctan(n))^(2n)

explain please! - Nov 28th 2010, 06:38 PMdwsmith
I first break it up. (n+1)^2=n^2+2n+1

Now you can evaluate three separate Summations.

I think the 2nd one can be evaluate like a geometric since ArcTan(Infinity)=pi/2 - Nov 28th 2010, 06:39 PMProve It
For the first, try the ratio test.

Evaluate $\displaystyle \displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$. If this is $\displaystyle \displaystyle <1$ the series converges absolutely, if this is $\displaystyle \displaystyle >1$ the series diverges. If this $\displaystyle \displaystyle =1$ the test is inconclusive. - Nov 28th 2010, 07:08 PMPlaythious
so for the first one I applied the ratio test and got

lim (as n goes to infinity) of (n^2 + 2n + 1) / ( (n+1)(n^2) ) + lim (as n goes to infinity) (2n + 1)/( (n+1)(n^2) + lim (as n goes to infinity) of 1/(n+1)

and since all of these go to zero then the sum converges absolutely?

and for the second one you have

sum (from 1 to infinity) 1 / (pi/2)^2n

and because pi/2 is greater than 1 it just diverges by geometric series test? - Nov 28th 2010, 07:14 PMdwsmith
You are mixing up two separate suggestions. I thought you wanted to solve the series but you just want to know if they converge. Just follow Post #3

$\displaystyle \displaystyle a_{n+1}=\frac{(n+2)^2}{(n+1)!}$

$\displaystyle \displaystyle \frac{\frac{(n+2)^2}{(n+1)!}}{\frac{(n+1)^2}{n!}}\ rightarrow \frac{(n+2)^2}{(n+1)^3}\rightarrow \frac{n^2}{n^3}\rightarrow\lim_{n\to\infty}\frac{1 }{n}=0$ - Nov 28th 2010, 07:19 PMPlaythious
ahh i see

so lim (as n goes to infinity) (n+2)^2 / ( (n+1)(n+1)2 which is just 0 so it converges abs

what about the second one? - Nov 28th 2010, 08:36 PMdwsmith
Try the nth root test