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Let $\alpha >0$, I want to study the convergence of the series $\sum_{n\geq 2}{\log(1+\frac{(-1)^n}{n^\alpha}})$. I know that the series $\sum_{n\geq 2}{\frac{(-1)^n}{n^\alpha}}$ is a convergent series by the alternating series test. Can we say that $\log(1+\frac{(-1)^n}{n^\alpha})$ is equivalent at $\infty$ to $\frac{(-1)^n}{n^\alpha}$ and then conclude that the series converges?

palio
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2 Answers2

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Note that for $\alpha > 0$ and $n \geq 2$,

$$ \log\left(1 + \frac{(-1)^{n}}{n^{\alpha}} \right) = \frac{(-1)^{n}}{n^{\alpha}} - \frac{1}{2n^{2\alpha}} + O(n^{-3\alpha}). $$

Thus if $\alpha \leq \frac{1}{2}$, then

$$ \sum_{n=2}^{N} \log\left(1 + \frac{(-1)^{n}}{n^{\alpha}} \right) = \sum_{n=2}^{N} \frac{(-1)^{n}}{n^{\alpha}} - \frac{1+o(1)}{2} \sum_{n=2}^{N} \frac{1}{n^{2\alpha}}. \tag{1} $$

diverges to $-\infty$. In fact, $\text{(1)}$ shows that this is an if-and-only-if condition:

$$ \sum_{n=2}^{\infty} \log\left(1 + \frac{(-1)^{n}}{n^{\alpha}} \right) \text{ converges if and only if } \alpha > \frac{1}{2}. $$

Sangchul Lee
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  • is it a big $o$ or a little $o$ used here? and how do you get from $\sum_{n=2}^N{O(\frac{1}{n^{3\alpha}})}$ to the part $\frac{-o(1)}{2}\sum_{n=2}^N{\frac{1}{n^{2\alpha}}}$ – palio Oct 27 '13 at 18:16
  • @palio,We have $$\sum_{n=2}^{N} n^{-s} = \begin{cases} \Theta(N^{1-s}), & s < 1 \ \log n + O(1), & s = 1 \ O(1), & s > 1. \end{cases}$$ That is, under the condition $\alpha \leq \frac{1}{2}$, the sum $\sum_{n=2}^{N} n^{-3\alpha}$ grows slower than $\sum_{n=2}^{N} n^{-2\alpha}$. This is why we can absorb the Big-Oh sum into a single small-oh. – Sangchul Lee Oct 27 '13 at 18:47
  • Please could you explain this, i worked hours on this without any result.. i already posted it here http://math.stackexchange.com/questions/561378/an-identity-with-big-o-and-little-o-notation – palio Nov 10 '13 at 18:34
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Your use of "equivalent" is not rigorous, and that turns out to be crucial in this problem. To use the Alternating Series Test, you need three things:

(i) $\ln(1+\frac{(-1)^n}{n^\alpha}) \to 0$ as $n \to \infty$.

(ii) The terms $\ln(1+\frac{(-1)^n}{n^\alpha})$ alternate in sign.

(iii) $|\ln(1+\frac{(-1)^{n+1}}{(n+1)^\alpha})| < |\ln(1+\frac{(-1)^{n}}{{n}^\alpha})|$ for all $n \geq 1$, or at least for large enough $n$.

(i) and (ii) are obvious. (iii) might take a little work. Maybe you could consider the cases of even $n$ and odd $n$ separately and us properties of logs (log of product is sum of logs, $\ln$ is increasing, etc.) I just tried the case where $n$ is even and the inequality in (iii) worked about to be true if $n$ was large enough (which is all you need).

EDIT: I just saw the answer above that concludes the series diverges. So I looked at (iii) again. (iii) turns out not to be true if $n$ is odd, for any $\alpha > 0$, even if you assume $n$ is large. So the Alternating Series Test tells you nothing.

Stefan Smith
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