# Infinite Series Problem

• Mar 16th 2010, 05:35 PM
mturner07
Infinite Series Problem
I'm sure this problem is obvious, but I don't know where to start.

Determine whether the series converges or diverges and give a good explanation how you arrived at your conclusion.

$\Sigma^{\infty}_{n = 1}n\sin{\frac{1}{n}}$

Also, for $\Sigma^{\infty}_{n = 1}\frac{1}{3 + n^2}$, I know I can use the integral test to show it converges, but can I also use the comparison test, comparing it to $\Sigma^{\infty}_{n = 1}\frac{1}{n^2}$?

• Mar 16th 2010, 05:53 PM
Krizalid
Quote:

Originally Posted by mturner07
$\Sigma^{\infty}_{n = 1}n\sin{\frac{1}{n}}$

the limit of the general term doesn't go to zero, so it diverges.

Quote:

Originally Posted by mturner07
Also, for $\Sigma^{\infty}_{n = 1}\frac{1}{3 + n^2}$, I know I can use the integral test to show it converges, but can I also use the comparison test, comparing it to $\Sigma^{\infty}_{n = 1}\frac{1}{n^2}$?

yes.
• Mar 16th 2010, 06:18 PM
mturner07
Isn't $\lim_{n \to \infty}n\sin{\frac{1}{n}} = 0$, since $\infty *\sin{0} = 0$?
• Mar 16th 2010, 06:40 PM
skeeter
Quote:

Originally Posted by mturner07
Isn't $\lim_{n \to \infty}n\sin{\frac{1}{n}} = 0$, since $\infty *\sin{0} = 0$?

$(\infty \cdot 0)$ is an indeterminate form.

$\lim_{n \to \infty} n \cdot \sin\left(\frac{1}{n}\right) = \lim_{\frac{1}{n} \to 0} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} = 1$
• Mar 16th 2010, 07:00 PM
mturner07
D'oh, thanks.