sigma 2^n +1/5^n +1
n=1

where n goes from 1 to infinity

This one was the easiest of all your series:
S_n = ∑2^n + ∑1/5^n + n
We are to find Lt(n->inf) S_n
If it tends to inf, we call it diverging, if it tends to a certain value, we call it converging!

This wasn't required though
∑1/5^n is small and hence negligible, but nonetheless it is a +ve quantity so, your answer will depend on ∑2^n + n
As you know ∑2^n + n is a diverging one as it tends to inf as n->inf

I am going to elaborate: ∑2^n = 2 + 2^2 + 2^3 + .....to inf
you know that a GP only converges when common ratio <1, but here c.r. = 2!

So, The series diverges!