First hint :
Second hint :
I need to find the sum of each series by computing their n-th partial sum and then taking the limit as n -> infinity.
1) Σ (infinity being on top, n=0 on bottom) 5/2^n - 4/3^n
I already took the limit by separating the two, and ending up with:
5/(1-(1/2)) - 4(1-(1/3)) = 4
I just need the n-th partial sum.
2) Σ (infinity being on top, n=1 on bottom) 1/(n^2-(1/4))
I really don't know how to do this one.