Why do I have a contradiction when using "Leibniz criterion" and " Integral test"

Question

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

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?

Answer 1

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 $(-1)^nf(n)$.

Answer 2

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.