1. ## 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.

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

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

2. Originally Posted by mturner07
$\displaystyle \Sigma^{\infty}_{n = 1}n\sin{\frac{1}{n}}$
the limit of the general term doesn't go to zero, so it diverges.

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

3. Isn't $\displaystyle \lim_{n \to \infty}n\sin{\frac{1}{n}} = 0$, since $\displaystyle \infty *\sin{0} = 0$?

4. Originally Posted by mturner07
Isn't $\displaystyle \lim_{n \to \infty}n\sin{\frac{1}{n}} = 0$, since $\displaystyle \infty *\sin{0} = 0$?
$\displaystyle (\infty \cdot 0)$ is an indeterminate form.

$\displaystyle \lim_{n \to \infty} n \cdot \sin\left(\frac{1}{n}\right) = \lim_{\frac{1}{n} \to 0} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} = 1$

5. D'oh, thanks.