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Does the series $\sum_{n=4}^\infty \frac{(-1)^n}{\log \log n}$ converge ?

I thought about alternating test, but for some reason this seems to easy. Why does it start with $n=4$? And how do I prove that $\frac1{\log \log n}$ is decreasing ?

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You could start it at another $n>1$ (So $\ln \ln n$ is defined.) – copper.hat Mar 6 '13 at 22:29
And why would "too easy" be a problem? $\log \log n$ is real for $n \ge 4$, right? – GEdgar Mar 6 '13 at 22:29
What is wrong with easy? – copper.hat Mar 6 '13 at 22:32
@GEdgar Yes, but normally they start with $n=1$, I thought there must be a reason... but maybe not – Kasper Mar 6 '13 at 22:32
log is increasing. We also need limit of our terms is $0$. Everything admittedly easy, but there are a few things to mention. And $\log 1=0$, so $\log\log 1$ is a problem. So is $\log\log 2$. They could have started at $3$. Or $17$. – André Nicolas Mar 6 '13 at 22:32

3 Answers 3

It is an alternating series of terms that diminish monotonically with limit 0, so it converges. It doesn't converge absolutely because its terms are larger than $1 /n$.

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With the derivative $(\log(\log x))'=\dfrac{1}{x\log(x)}>0$ so the function is increasing for $x>1$, so the sequence $\dfrac{1}{\log(\log n)}$ is decreasing to $0$ and the alternating test is applicable.

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It really is that easy. If $f(x)$ is increasing, then $\frac{1}{f(x)}$ is decreasing. The denominator of the sequence is increasing. You can check using the first derivative test that $\frac{d}{dx} \ln( \ln x) >0$ (Note: the sequence log(log(n)) is increasing if the function log(logx)) is increasing). Thus, the original sequence is increasing. Alternatively, you can use induction.

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@JonasMeyer Sorry. Changed iff to if. Thanks. – BCLC Sep 25 '14 at 12:26

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