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How do you prove that the series $ \sum_{n \geq 0} \frac{(-1)^n}{n^{1+1/n}}$ converges? I have shown (simple ratio test) that if it's not alternate it diverges. Any ideas?

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    OK. So it's an alternating series. Any idea of what tests work on those? – 2017-01-01
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    The ratio test does not show anything here. Also, saying "if it's not alternate it diverges" doesn't really make sense. – 2017-01-01

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Hint: One way to approach this is the alternating series test. We would need to show $1/n^{1+1/n} \to 0$ (easy) and that the terms $1/n^{1+1/n}$ are eventually decreasing (harder).

Here's another way: Show

$$\tag 1 \sum_{n=1}^{\infty}\left (\frac{1}{n}-\frac{1}{n^{1+1/n}}\right ) < \infty.$$

If we have that, we can write

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^{1+1/n}} = \sum_{n=1}^{\infty}(-1)^n\left (\frac{1}{n^{1+1/n}}- \frac{1}{n}\right ) + \sum_{n=1}^{\infty}\frac{(-1)^n}{n}.$$

Thus our series is the sum of two series, the first of which converges absolutely by $(1),$ the second of which converges by the alternating series test. Hence our series converges.

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    May as well give this to you. :) – 2017-01-01
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    You could edit your answer by handling the "decreasing" part of it; then we'd have two proofs. – 2017-01-01
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    I could, and its obviously not very hard, but I happen to be busy atm and you have a wonderful answer already. – 2017-01-01
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    How would you show (1) converges? – 2017-01-01
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    The $n$th term of $(1)$ is less than $$\frac{n^{1/n}-1}{n}.$$ Do a little work to see $n^{1/n}-1 = O((\ln n)/n)).$ So $(1)$ converges if $\sum_n (\ln n)/n^2$ converges, which it does. – 2017-01-02