# Why do I have a contradiction when using “Leibniz criterion” and “ Integral test”

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

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?

• 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. – ajotatxe Oct 9 '18 at 20:32
• Note, you have proved that the series is conditionally convergent – qbert Oct 9 '18 at 21:47

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

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 .