Let's consider the series $$s(x)=\sum_{n=2}^\infty \left(\frac{1}{n}-\frac{x}{\ln n}\right).$$
I want to prove it diverges for all $x$. To do that I want to compare to some harmonic series but I don't know how to start. Any idea about how to do that?
In a more particular case, I'm interested when $x=1$.
I tried to express it like this this:
$$s(x)=\sum_{n=2}^\infty \frac{\ln n- nx}{n\ln n} $$ but it doesn't seem to simplify the problem.
On
Note that for $x=1$
$$\sum_{n=2}^\infty \left(\frac{1}{n}-\frac{1}{\ln n}\right)=-\sum_{n=2}^\infty \left(\frac{1}{\ln n}-\frac{1}{n}\right) \quad f(n)=\frac{1}{\ln n}-\frac{1}{n}>0$$
then by Cauchy condensation test
$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$
$$\sum_{n=2}^\infty \left(\frac{1}{\ln n}-\frac{1}{n}\right)\ge\frac12\sum_{n=1}^\infty \left(\frac{2^n}{\ln 2^n}-\frac{2^n}{2^n}\right)=\frac12\sum_{n=1}^\infty \left(\frac{2^n}{n\ln 2}-1\right)=\infty$$
For the general case note that eventually $\forall x\in \mathbb{R^+}$ (the case $x\le0$ is trivial)
$$\sum_{n=2}^\infty \left(\frac{1}{n}-\frac{x}{\ln n}\right)=-\sum_{n=2}^\infty \left(\frac{x}{\ln n}-\frac{1}{n}\right) \quad f(n)=\frac{x}{\ln n}-\frac{1}{n}>0$$
then apply the condensation test.
The case $x\le0$ is easy, since each term in the series will exceed $\frac1n$. For the case $x>0$, the terms in your series are eventually all negative (why?) so you should prove that $$ \sum \left(\frac x{\ln n} - \frac1n\right) $$ diverges to $\infty$. To prove that, argue that $$ \frac x{\ln n} - \frac1n>\frac1n $$ for all $n$ sufficiently large (exactly how large depends on $x$), which follows from $$ \lim_{n\to\infty}\frac{\ln n}n = 0. $$