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.

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Expressions in the exponents need to be enclosed by { and }, not parentheses. Also, you should include the actual question in the body of the message, don't realy exclusively on the title. – Arturo Magidin Nov 4 '10 at 19:29
You are misunderstanding the Alternating Series Test. The test asks you to check what happens to the terms, not what happens to the series. – Arturo Magidin Nov 4 '10 at 19:46
+1 for explaining what you've done and where you are stuck. – Qiaochu Yuan Nov 4 '10 at 20:08
+1 for the same reason as Qiaochu. Also, if you click the "edit" link you can see how Arturo fixed the exponent. – AD. Nov 4 '10 at 21:01

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.

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I'm surprised you haven't gone deaf from all those expressions "screaming" at you. :) – J. M. Nov 4 '10 at 22:43
Thanks. Your advice is always incredibly helpful and you always find the errors in the things I forget ie series needs to be positive for large n to apply some tests etc. – fdart17 Nov 4 '10 at 22:47
Oh. I just realized. I also have to show that the series is not absolutely convergent. I worded my question very poorly, since it asks weather it diverges, converges, or absolutely converges. But now it looks like a matter of showing it's absolutely convergent. I will try some tests and post results. – fdart17 Nov 4 '10 at 23:13
At the moment I have $lim 2(n^{(2/n)-1})$ from using l'hospitals once. Anything we can do with this? – fdart17 Nov 4 '10 at 23:20
Awesome by the root test the series is absolutely convergent. – fdart17 Nov 4 '10 at 23:27

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 $f(x)=\sin(x)$.

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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.

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Yes the Alternating Test will work.

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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.]

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No, you also need $2^{1/(n+1)}-1\le 2^{1/n}-1$ for all – Robin Chapman Nov 4 '10 at 19:34
@ACW: You need to show both that the limit is zero, and that the sequence is eventually decreasing. – Arturo Magidin Nov 4 '10 at 19:34
large enough $n$. – Robin Chapman Nov 4 '10 at 19:34