Prove that $\sum\limits_{n=1}^{\infty}(-1)^n(2^{1/n} - 1)$ is convergent 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!
 A: 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 $f(x)=\sin(x)$. 
A: In many cases expanding the exponential works, in this case:
$2^{\frac{1}{n}} = \exp ( \frac{\ln 2}{n} ) = 1 + \frac{\ln 2}{n} + c_n $. Where $ \lim_{n \to \infty} n^2 c_n  =  (\ln 2)^2 \times ( \exp ( \frac{\ln 2}{n} ) - 1 - \frac{\ln 2}{n} ) /( \frac{\ln 2}{n})^2 = \frac{(\ln 2)^2}{2}.$ So, $c_n$ converges absolutely.
Now, $\sum_{ n \geq 1} (-1)^n (2^{\frac{1}{n}} - 1) = \sum_{n \geq 1} (-1)^n \frac{\ln 2}{n} + \sum_{n \geq 1} (-1)^n c_n$, which clearly converges.
A: This is an alternating series, so it suggests (nay, screams) for the use of the Alternating Series Test. You need to show that $\lim\limits_{n\to\infty}(2^{1/n}-1) = 0$, and that for all sufficiently large $n$, $2^{\frac{1}{n+1}}-1 \lt 2^{1/n} - 1$. 
The latter is equivalent to showing that $2^{\frac{1}{n+1}}\leq 2^{\frac{1}{n}}$; you might want to consider taking logarithms, say base $2$, which respect the order (that is, if $0\lt a\lt b$, then $\log_2(a)\lt\log_2(b)$). As to the former, it amounts to showing that $\lim\limits_{n\to\infty}2^{1/n} = 1$. Since $a^b = e^{b\ln(a)}$, this is equivalent to showing that $\lim\limits_{n\to\infty}e^{\frac{\ln(2)}{n}} = 1$; and since the exponential is continuous, this amounts to checking what happens to the exponent as $n\to\infty$. 
Note also that none of the other tests that you mention are even applicable! The integral test can only be applied to a series $\sum a_n$ in which all terms $a_n$ are nonnegative (which is not the case here), and where you can find a positive, decreasing, and continuous function $f(x)$ such that $f(n)=a_n$ for all $n$ (or at least all sufficiently large $n$). So it is not applicable here. The Comparison Test requires the series to have positive terms, as does the Limit Comparison Test. You could combine them with the Absolute Convergence Test by checking the convergence of $\sum|a_n|$, and then deducing the convergence of $\sum a_n$ if the series of absolute values converges, and apply one of these tests to the series $\sum|a_n|$. But you cannot apply any of the Integral, Comparison, or Limit Comparison Tests to this series directly.
A: I believe that the alternating series test is the right way to go.  All you have to do is prove that $\lim (2^{1 \over n} - 1)$ is zero.  That is, to employ the test, you don't need to show that the limit of the series is zero, but merely that the limit of the summands is zero.
[ETA: Arturo reminds us that one must also show that the summands strictly decrease.  But the AST is still the correct approach.]
