Series: comparison test

• Sep 13th 2009, 10:14 AM
Apprentice123
Series: comparison test
Use the comparison test to investigate the convergence or divergence

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

My solution:

p-series $\frac{1}{n^2}$ converges

$\frac{1}{n^2+1} < \frac{1}{n^2}$

then $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ converges

Is correct ?

2) $\sum_{n=1}^{\infty} \frac{n!}{\sqrt{n(n+1)}}$

$n!$ diverges in the ratio test, then I use the $\lim_{n ->{\infty}} \frac{\frac{n!}{\sqrt{n(n+1)}}}{n!} = 0$

But is 0, how can I determine the convergence or divergence?
• Sep 13th 2009, 02:38 PM
Apprentice123
In problem 2:

$b_n = \frac{n!}{n}$ In the ratio test:

$\lim_{n \to \infty} \frac{(n+1)n!}{n+1}. \frac{n+1}{n!} = \infty$ Diverges

Then, $\frac{n!}{n} > \frac{n!}{\sqrt{n(n+1)}}$

$\sum_{n=1}^{\infty} \frac{n!}{\sqrt{n(n+1)}}$ => Diverges

It is correct ?

3) $\sum_{n=0}^{ \infty} \frac{1}{n!}$

My solution:
$b_n = \frac{n}{n!}$ In the Ratio test Converges

$\frac{1}{n!} < \frac{n}{n!}$
Then:

$\sum_{n=0}^{ \infty} \frac{1}{n!}$ is convergence

Is correct ?

4) $\sum_{n=1}^{\infty} \frac{(n+1)}{n^3}$
How can I resolve this?