# Convergence and divergence, two questions

• May 15th 2009, 07:55 AM
enormousface
Convergence and divergence, two questions
hi y'all, here is a two part question that i just cant figure out. I've been staring at them for a while no, and i think that the firs tone is not convergent and the second is convergent, but i dont know how to go about proving that (or confirming if i'm right/wrong) Please let me know what you think.

do the following series converge or diverge? Whatever you put, prove you answer.

a.) [epsilon]infinity,n=1 (e^(1/n))/(n^2)

b.) [epsilon]infinity,n=1 (n!)/(2*4*6. . . . . .(2n))
• May 15th 2009, 08:31 AM
Calculus26
They both converge--See attachment
• May 15th 2009, 08:44 AM
chisigma
a) $\forall n$ is $e^{\frac{1}{n}}<3$ so that is

$\frac{e^{\frac{1}{n}}}{n^{2}} < \frac{3}{n^{2}}$

The series...

$\sum_{n=1}^{\infty} \frac{3}{n^{2}}$

... converges so that also converges the series...

$\sum_{n=1}^{\infty} \frac{e^{\frac{1}{n}}}{n^{2}}$

b) is...

$\frac{1\cdot 2 \cdot 3 \dots (n-1)\cdot n}{2\cdot 4\cdot 6 \dots (2n-2)\cdot 2n} = \frac{1}{2^{n}}$

... so that...

$\sum_{n=1}^{\infty} \frac{1\cdot 2 \cdot 3 \dots (n-1)\cdot n}{2\cdot 4\cdot 6 \dots (2n-2)\cdot 2n} = \sum_{n=1}^{\infty} \frac {1}{2^{n}} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$

Kind regards

$\chi$ $\sigma$