Determine the convergence/divergence of this series with all possible tests

[mathnoobie]

Homework Statement

series from n=0 to infinity of
(n+1)/3^n

Did I do this correctly? Also, I'm pretty positive that I can use the comparison test/limit comparison test, but I can't seem to figure out what to compare it to, could someone shed a little light? For the integral test, the integral is quite intimidating and I don't really know where to start, could someone suggest what integration technique to try?

relevant equations:
Ratio Test

The Attempt at a Solution

so first I used the ratio test
lim n->infinity |((n+2)/3^(n+1)) * ((3^n)/(n+1))|
= lim n->infinity (n+2)/(3(n+1))
dividing by n/n
you have the limit = 1/3<1, therefore it converges by the ratio test.

 

[micromass]

That's good.

If you want to do the comparison test, then maybe apply

(n+1)&lt;2^n

for large n?

 

[mathnoobie]

That's good.

If you want to do the comparison test, then maybe apply

(n+1)&lt;2^n

for large n?

So then I would have a convergent geometric series since 2/3<1.
Wow, beautiful. I would've never thought of that on my own, been staring at it for 30minutes trying to compare it to random things.

 

[johnqwertyful]

You could simply divide and you would have two geometric series.

Also trivially, it's probably 1 to infinity, because of the n^3 on the bottom.

 

[mathnoobie]

You could simply divide and you would have two geometric series.

Also trivially, it's probably 1 to infinity, because of the n^3 on the bottom.

I'm confused on where the n^3 came from. Was that a typo and you meant 3^n?

 

[Mark44]

You could simply divide and you would have two geometric series.

Also trivially, it's probably 1 to infinity, because of the n^3 on the bottom.

I'm confused on where the n^3 came from. Was that a typo and you meant 3^n?

I think that johnqwertyful misread your post.

 

[johnqwertyful]

Oops, misread.