Prove whether or not the series converges? $\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{1/n}}$ Any help is much appreciated on this!
 A: 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.
A: When you see $$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{1/n}}$$
you may not immediately think about using the Divergence Test, though in general, that should be the first test you try on a given series.
I think the most natural approach would be to use the Alternating Series (Leibniz Test) here though.
This test states that if $b_n$ is positive, decreasing, and $b_n \to 0$ as $n\to\infty$, then our series converges. $b_n$ is the positive part of the summand $(\frac{1}{n^{1/n}})$.
It is easy to see that $b_n$ is positive for $n \in [2, \infty)$. Now, look at the limit of $b_n$ as $n \to \infty$:
$$\lim_{n\to\infty} \frac{1}{n^{1/n}} = 1 \ne 0\implies \sum{a_n} \quad \text{diverges}$$
However, as noted by Jonas Meyer in the comments, the terms of $b_n$ are in fact increasing, so it is immediate that the series diverges because the hypotheses for the AST are not met.
