Q:

What's the series and what's the radius of convergence of this (power) series?

Find the convergence radiuses of this power series:
$1 + n + n^{4} + n^{9} + n^{16} + n^{25} + n^{36} + ...$

First of all, I'm surprised it says $radiuses$ instead of $radius$. I know you find the radius by finding the limit of the series aka where it converges and take its reciprocal. But how and is it even possible to find more than just 1 radius for a power series? That would be my first question.
For the task, it doesn't really look like a series because the totals formula is missing, instead we got "...". So I would change it to:
$\sum_{k=0}^{\infty } n^{k^{2}}$
Now we need to find out if this series converges and what its limit point towards $\infty$ is.
I don't know what rule to use here, till now we mostly had fractions where you could easily use the ratio test.
However, as it looks like this series will go to $\infty$, so it will diverge thus there is no limit point and no convergence radius...?
Can't be or maybe I converted it badly to a series at the beginning?

A:

As @Dr.MV said, we can use the comparison test to solve this problem.

$\lvert n \rvert < 1 \rightarrow$ In this case, we have $\lvert n^{k^2} \rvert < \lvert n^k \rvert$, so since $\sum_{k=0}^\infty n^k$ converges, so does $\sum_{k=0}^\infty n^{k^2}$.
$\lvert n \rvert \geq 1 \rightarrow$ In this case, we have $\lvert n^{k^2} \rvert \geq \lvert n^k \rvert$, so since $\sum_{k=0}^\infty n^k$ diverges, so does $\sum_{k=0}^\infty n^{k^2}$.

Thus, the sum converges if and only if $\lvert n \rvert < 1$.

