How to prove that the second condition for Leibniz test is met for the series?

Question

So, here is the series: $\sum^{\infty}_{2}\frac{k}{(k\ln x +x^2)^2}$. I need to show that $\frac{k+1}{((k+1)\ln x +x^2)^2} - \frac{k}{(k\ln x +x^2)^2} \ge 0 \ \ \ \forall x\in (0, \infty), \forall k \ge 2 $

How to do that?

Answer 1

HINT:

As written, the series diverges for all $x$. We shall assume that the series of interest is $\sum_{k=2}^\infty \frac{(-1)^k k}{(k\log(x)+x^2)^2}$.

Then, we can write the positive part of the general terms of the alternating series as

$$\begin{align} \frac{k}{(k\log(x)+x^2)^2}&=\frac1{\log(x)}\left(\frac{k\log(x)+x^2-x^2}{(k\log(x)+x^2)^2}\right)\\\\ &=\frac1{\log(x)}\left(\frac1{k\log(x)+x^2}-\frac{x^2}{(k\log(x)+x^2)^2}\right) \end{align}$$

Note that we have split the original series into the difference of two series, both of which have positive parts that are monotonically decreasing by inspection.

Can you finish now?