Series convergence problem I´m having trouble analyzing the convergence of another series:
$$\sum_{n=1}^\infty (-1)^n \frac 1 {n^{\frac 1 n}}$$
If you could just give me a hint about which test for convergence should I use, it would be great.
Thanks so much!
 A: Hint: We have $n\lt 2^n$, so $n^{1/n}\lt (2^n)^{1/n}=2$.
A: An essential limit to know is that:
$$\sqrt[n]{n}\to 1$$
Which is a consequence of the following also essential limit:
$$(\forall a>1)\quad \frac{a^n}{n}\to0$$
The latter can be proven by looking at ratios of successive terms, you'll find that it's almost like a geometric sequence, for large enough $n$. In particular, then, we have eventually $n<a^n$, and so $\sqrt[n]{n}<a$, and this is for all $a>1$. Since $\sqrt[n]{n}$ is clearly eventually greater than $1$, we have $\sqrt[n]{n}\to1$. Incidentally, by shuffling exponents around and taking the limit of a product, this means that $\sqrt[n]{n^k}\to1$, too, and without too much more work, even that $\sqrt[n]{P(n)}\to1$ for any polynomial $P$.
Anyway, this almost immediately implies that the general term of your sequence converges to $1$ (in absolute value), and thus that the series does not converge.
A: The series diverges because
$$
\lim_n\frac{1}{n^{1/n}}=\lim_n\exp(-\frac{\ln n}{n})=1 \ne 0.
$$
