# Math Help - Series convergance?

1. ## Series convergance?

does these series conv. or diver.?

1. to infiity n=1 $(arctan(n))/(n^3)$
im not sure where to start exactly, i was thinking Ratio or root test but im lost

2. to infinity n=1 $((-3)^n-1) /(4^n)
$

same as above, not sure

any help/tips would be greatly appreciated

Thanks

2. 1. arctan(n)<pi/2

So use comparison test with the conv pseries 1/n^3

2. rewrite (-3/4)^n - (1/4)^n the sum of 2 convergent geometric series

3. Originally Posted by Itsygo
does these series conv. or diver.?

1. to infiity n=1 $(arctan(n))/(n^3)$
im not sure where to start exactly, i was thinking Ratio or root test but im lost

2. to infinity n=1 $((-3)^n-1) /(4^n)
$

same as above, not sure

any help/tips would be greatly appreciated

Thanks
I'll have a think about these, but for 1., I would say you could use a comparison test (could someone verify?)

Since $-\frac{\pi}{2}\leq\arctan{n}\leq\frac{\pi}{2}$ we can say that

$\sum_{n = 1}^{\infty}{\frac{\arctan{n}}{n^3}} \leq \frac{\pi}{2}\sum_{n = 1}^{\infty}{\frac{1}{n^3}}$, which is a p-series.

This p-series converges to a number, and since the original series is less than this number, the original series must also converge.