Divergent:

\sum (n=1 to infinity) [4 + 3^n] / (2^n) = \sum (n=1 to infinity) 4/ (2^n) + \sum (n=1 to infinity) 3^n / 2^n

................= \sum (n=1 to infinity) 4/ (2^n) + \sum (n=1 to infinity) (3/2)^n

The first of the sums in the above converges as its a geometric series with factor <1, while the second sum diverges as again it is a geometric series with factor >1.

RonL