# Need help with these convergence Tests?

• November 28th 2010, 06:35 PM
Playthious
Need 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)

• November 28th 2010, 06:38 PM
dwsmith
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
• November 28th 2010, 06:39 PM
Prove It
For the first, try the ratio test.

Evaluate $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$. If this is $\displaystyle <1$ the series converges absolutely, if this is $\displaystyle >1$ the series diverges. If this $\displaystyle =1$ the test is inconclusive.
• November 28th 2010, 07:08 PM
Playthious
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?
• November 28th 2010, 07:14 PM
dwsmith
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

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

$\displaystyle \frac{\frac{(n+2)^2}{(n+1)!}}{\frac{(n+1)^2}{n!}}\ rightarrow \frac{(n+2)^2}{(n+1)^3}\rightarrow \frac{n^2}{n^3}\rightarrow\lim_{n\to\infty}\frac{1 }{n}=0$
• November 28th 2010, 07:19 PM
Playthious
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