Prove whether or not the series converges? $\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{1/n}}$ Any help is much appreciated on this!

You can use the Test for Divergence here. For a series to be convergent you need the limit of the terms to go to zero, i.e. that
$$
\lim_{n\to \infty} \frac{(-1)^n}{n^{1/n}} = 0.
$$
That is is equivalent to needing that
$$
\lim_{n\to \infty} n^{1/n} = \pm\infty.
$$
So if this limit is not infinity (or negative infinity), then the series is divergent.
However by noting that with $y = n^{1/n}$ you have $\ln(y) = \frac{\ln(n)}{n}$. You can probably show that the limit of $\ln(y)$ as $n\to \infty$ is $0$. So that means that
$$
\lim_{n\to \infty} n^{1/n} =\lim_{n\to \infty} y = 
\lim_{n\to \infty} e^{\ln(y)} = e^{\lim_{n\to \infty} \ln(y)}  = e^0 = 1.
$$
So this series is divergent.
