# series convergence

• Feb 1st 2013, 10:24 AM
randommathnoob
series convergence
Hi i have following two series:

1.
$\sum_{ 1}^{\infty } \sin(\frac{1}{n})\cos(\frac{1}{n^{2}}})$

2.
$\sum_{ 1}^{\infty } \frac{1}{n^{3}}\sin(\frac{-1^{n}}{ \sqrt{n} })$

and i dont really know how to touch them.
I'd be really greatful for any kind of help,
Cheers.
• Feb 1st 2013, 05:25 PM
hollywood
Re: series convergence
I assume you want to know if they converge or not. For number 2, you would use the comparison test (compare to $\frac{1}{n^3}$).

Number 1 is pretty tricky. I'm pretty sure it diverges - as n goes to infinity, $\sin\frac{1}{n}$ looks like $\frac{1}{n}$ and $\cos\frac{1}{n^2}$ looks like 1, so your series looks like a harmonic series. But that's not a proof - that's just intuition.

If you use $\sin{x}>x-\frac{x^3}{6}$ and $\cos{x}>1-\frac{x^2}{2}$, then I think you can prove divergence. The series you compare against will be the harmonic series with some convergent series added or subtracted. They don't affect the divergence.

You might have heard of the limit comparison test - you should check if this applies.

Let us know how it goes.

- Hollywood
• Feb 2nd 2013, 12:45 AM
randommathnoob
Re: series convergence
I've managed to do both of them using the limit comparison test,
for the first one we can simply plug

$\sum_{ 1}^{\infty } \frac{1}{n}$

and for the second one i've used absolute convergence + limit comparison test with.

$\sum_{ 1}^{\infty } \frac{1}{ n^{3}\sqrt{n} }$
• Feb 2nd 2013, 07:56 AM
hollywood
Re: series convergence
I believe that's correct.

- Hollywood