How do you find the sum of 2 ^k/2 when k =0 and goes to infinity, and does it converge or diverge.?

Question

I know its a geometric series but the ^k/2 is whats getting me. If it were to find the sum of 2^k I would be good.lol

Answer 1

2^(k/2) = [sqrt(2)]^k. Since sqrt(2) > 1, the infinite series diverges.

Answer 2

Treat it as the sum of two separate series, one of the form 2^n and one of the form 2^(n+0.5). The latter is subject to an obvious simplification (can you see what it is?), and then you can deal with the two series separately (actually, the same series, twice). If two series both converge, their sum is convergent; otherwise, the sum is divergent. In this case, both diverge. The more interesting case arises if you are dealing with 2^(-k/2), which converges.

Answer 3

This is a geometric serie with ratio 2^1/2 = sqrt2.
As the ratio is greater than 1 it diverges

The sum to the kth term would be (2^k/2 -1)/(2^1/2 -1)