Determine wheter the series is convergent or divergent?

Question

Find the sum if Convergent:

∑ where n =1 to infinity (3/(n(n+3)) + 5/4^n)

The first it's 3 (I think) but I can't get the second geometric series right. What I do is:

5∑(1/4)^n, and then 5∑(1/4)(1/4)^n-1, so a = 5/4, r = 1/4, it's sum would be: s = a/1-r, but I get 5/3, when I add up that to the 1 from the other series I get 8/3.

The answer is:7/2 :( What am I doing wrong? Thanks!

Oops, I meant "whether" >_> And the first series looks wrong ;(

Answer 1

The geometric starts with 1/4 so its sum would be
5/4(1/(1-1/4) =5/3
The first is
1/n-1/(n+1)+1(/n+1)-1/(n+2)+1/(n+2)-1/(n+3)
Remember that you start summing from n=1
so this sum is
1+1/2+1/3 =11/6
and+10/6 you get 21/6 = 7/2