TITLE: Prove that $\sum\limits_{n=1}^{\infty}(-1)^n(2^{1/n} - 1)$ is convergent
QUESTION [9 upvotes]: I want to prove that the series
$$ \sum_{n=1}^{\infty}(-1)^n\left(2^{1/n}-1\right)$$
converges.
I am fairly certain that this converges.  Using the ratio or root test does not seem to work.  Therefore the other tests I am left with are alternating series test, intergral test, comparison test and limit comparison test (that we know about).  
Seems like it would be a messy integral and that would not work or I would not be able to solve it.
I think I could show this using the alternating series test IF the limit of the series is 0.  My problem is formally showing that the limit of this series is 0.  I can't quite get past that.
My only option left if the alternating series test does not work is the limit comparison test, since the series is not > 0 for large n, and therefore we cannot use the comparison test.  If this is the case I honestly don't have a clue what other series I should compare this with.  
Thank you for reading!

REPLY [3 votes]: A general suggestion: There are a few problems asking for convergence of divergence of series of the form $\sum_n (-1)^ng(n)$ or $\sum_n g(n)$ for some function $g$ for which one can easily rewrite $g(n)$ as
$$f(1/n)-f(0)$$ for some $f$ differentiable at 0.
In all these cases one can try to replace $g(n)=f(1/n)-f(0)$ with $f'(0)/n$ and use elementary estimates and inequalities for the remainder error, if needed (sometimes a limit comparison test will already suffice).
For example, in the problem here, $g(n)=2^{1/n}-1$ so blatantly we can try $f(x)=2^x$. We have $f'(0)=\ln 2$, which is the estimate that appears in acarchau's answer.
For another example, $\sum_n\sin(1/n)$ diverges: Take $f(x)=\sin(x)$.