Does a series always diverge if its sequence isn't a null-sequence?

Question

I have the following series:

$$\sum_{n = 1}^\infty 2^{n^2}z^n$$

The task is to give its radius of convergence. I solved that one using the root-test and came to the same answer. But the solution states the following:

The radius of convergence is $0$, because the sequence $(2^{n^2}z^n)_{n\in\mathbb{N}}$ isn't a null-sequence.
Proof: It is clear, that there is an $N \in \mathbb{N}$, so that $2^nz > 1 \forall n\ge N$. Therefore we conclude:

$$2n^2z^n = (2^nz)^n > 1 \forall n \ge N$$

Now that proof is clear to me. I just wanted to know:
Is this reasoning "not a null-sequence $\Rightarrow$ series diverges" always possible? I have never seenthat approach before...

Thank you very much for your help!
FunkyPeanut

Answer 1

The series $\displaystyle\sum_{n\ge0} a_n$ is convergent if and only if the sequence of partial sum $(S_n)_n =\left(\displaystyle\sum_{k=0}^n a_k\right)_n$ is convergent and in this case $$\lim_{n\to\infty}S_{n}-S_{n-1}=\lim_{n\to\infty}a_n=0$$ and simply we get the desired result by contrapositive.