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$$\sum \frac{(-1)^{n}}{n\cdot \ln n}$$

leibniz criterion: The series $\frac{1}{n\cdot \ln n}$ is monotonically decreasing. Also $\lim_{n \to \infty}\frac{1}{n\cdot \ln n}=0$

Thus, the the series is convergent.

however, these criteria also met with "integral test".

Using Integral test: $\int \frac{1}{x\cdot \ln x}dx$ using the formula $\int \frac{f'(x)}{f(x)}dx=\ln (f(x))$

I get: $\ln (\ln x)$

solving the latter with "infinity" as an upper bound gets me a result of "Divergent" which for some reason contradicts Leibniz criterion that gets me an answer of "Convergent"

why is that?

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  • $\begingroup$ You have proved that $\sum\frac{(-1)^n}{n\ln n}$ is convergent and that $\sum\frac1{n\ln n}$ is divergent. No contradiction so far. $\endgroup$ – ajotatxe Oct 9 '18 at 20:32
  • $\begingroup$ Note, you have proved that the series is conditionally convergent $\endgroup$ – qbert Oct 9 '18 at 21:47
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The integral test relates $\sum f(n)$ and $\int f(x)$ where $f$ is a non-negative function.

In your case, you've selected $f(x)=\frac{1}{x\ln x}$, but the terms of your series are not given by $f(n)$. They're given by $(-1)^nf(n)$.

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The series of interest is $\sum_{n\ge 2}^\infty \frac{(-1)^n}{n\log(n)}$, not $\sum_{n\ge 2}^\infty \frac{1}{n\log(n)}$.

By applying Leibniz's test, you showed correctly that the alternating series, $\sum_{n\ge 2}^\infty \frac{(-1)^n}{n\log(n)}$, converges.

And by using the integral test on the series $\sum_{n\ge 2}^\infty \frac{1}{n\log(n)}=\sum_{n\ge 2}^\infty \left|\frac{(-1)^n}{n\log(n)}\right|$, you showed correctly that the series, $\sum_{n\ge 2}^\infty \frac{(-1)^n}{n\log(n)}$, does not converge absolutely .

There is no contradiction here.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Oct 10 '18 at 14:28

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