What test would I use to determine the convergence of the series below?

Question

anonymous · Answer

$$\sum_{n = 1}^{\infty} \sin(1/n)/n^2$$

zarkon · Answer

comparison

anonymous · Answer

Comparing it to what?

anonymous · Answer

Would the integral test be valid in this situation?

abb0t · Answer

comparison test with $\large \frac{1}{n}$

anonymous · Answer

1/n is greater?

abb0t · Answer

$\large \frac{1}{n^2}$

anonymous · Answer

I'm probably missing something, but wouldn't sin(1/n) make the original series larger?

anonymous · Answer

Wait, you're right I tried plugging things in and it worked

abb0t · Answer

you could do a limit comparison test for this, also. But your focusing on $\frac{1}{n}$

anonymous · Answer

So it would diverge?

zarkon · Answer

no

abb0t · Answer

$\frac{1}{n}$ diverges.

anonymous · Answer

If you're using the comparison test with 1/n then both of them diverge

zarkon · Answer

$$\frac{1}{n^2}$$ converges

zarkon · Answer

$$\left|\frac{\sin\left(\frac{1}{n}\right)}{n^2}\right|\le\left|\frac{1}{n^2}\right|$$

anonymous · Answer

ok thanks, you got 1/n^2 because it's the same denominator?

anonymous · Answer

Since$$-1\le\sin\left(\frac{1}{n}\right) \le 1$$ for all n, dividing everything by n^2 yields:$$\frac{\sin\left(\frac{1}{n}\right)}{n^2}\le \frac{1}{n^2}$$

zarkon · Answer

you could have also used the integral test as Joe had mentioned

anonymous · Answer

ok thanks!