Determine whether the series converges or diverges: Sum [(n+3^n)/(4^n)]; n=0 to inf. I am using the Ratio test, but is stuck

Question

experimentx · Answer

(n+3^n)/(4^n)

n/4^n + (3/4)^n <---Geometric progression ... converges since 3/4 < 1

just have to prove n/4^n converges,  must be easy.

experimentx · Answer

looks like ratio test will easily determine http://www.wolframalpha.com/input/?i=lim+n-%3Einf+%28%28n%2B1%29%2F4%5E%28n%2B1%29%29%2F%28n%2F4%5En%29

anonymous · Answer

@experimentX way is cleaner but from the ratio test we get:$$\frac{(n+1)3^{n+1}}{4^{n+1}} \frac{4^n}{n3^n} = \frac{3}{4} \frac{n+1}{n}$$Limit of (n+1)/n = 1, so we have 3/4 < 1, which is true.

experimentx · Answer

yeah ... you are right. but i thought that was cleaner.

anonymous · Answer

Yes, I got to lim |n+4/4n| = 1/4, converges. 
I tried to use geometric, but I didnt know how to prove if (n/4^n) is conv or div.

anonymous · Answer

Thanks!! You shall see more!

anonymous · Answer

Ratio test should work for that also :-)

anonymous · Answer

Oooh, okizzle!