# How is $\frac{1}{x(\ln x)^2}$ decreasing?

I am new to integral testing, so my professor is trying to explain how to use the test on this series from 2 to infinity, and it already satisfies the 1st condition by saying "it is obviously decreasing" and moving on to evaluate. How is it obviously decreasing?
I differentiated the function to get $$\dfrac{2+\ln x}{x^2 (\ln x)^3}$$, but still couldn't easily spot where is it below $$0$$. So is it right to say that any $$x$$ value below $$1$$ will make $$f'(x)$$ negative, and this is where $$f(x)$$ is decreasing? I plotted it on the graph and it looks like, after $$x=1$$, $$f(x)$$ drops down and starts decreasing, so I don't know what to think anymore. My understanding of this concept to test for convergence or divergence is hazy, so I would really appreciate if someone could provide clarity.

edit: iam very sorry i forgot there is minus sign for f '(x) so i clearly just confused myself for no reason. apologies!

• Do you know that $x$ and $(\ln x)^2$ are both increasing on $(1,\infty)$? – Umberto P. Mar 29 at 15:49
• Is your function $\frac{1}{x\ln(x)^2}$ or $\frac{1}{x}\ln(x)^2$? – Bernard Masse Mar 29 at 15:57
• @BernardMassé its the 1st one but (lnx)^2 not x^2 – Aya Rahima Mar 29 at 15:58
• @UmbertoP. yes i know. could you please tell me how will that help ? iam little confused – Aya Rahima Mar 29 at 16:00
• Since both are increasing so is their product. What about the reciprocal of an increasing function? – Umberto P. Mar 29 at 16:02

For any $$g$$, if $$g(x)$$ is increasing, then $$\frac{1}{g(x)}$$ is decreasing. Since $$x$$ is $$(\ln x)^2$$ are both increasing for $$x>1$$, $$x(\ln x)^2$$ is increasing and thus $$\frac{1}{x(\ln x)^2}$$ is decreasing.
For example: it is the product of two (positive) descending sequences ($$\;x\ge 2\;$$), or also:
$$f(x):=\frac1{x\log^2x}\implies f'(x)=-\frac{\log x+2}{x^2\log^3x}<0$$
for any $$\;x\ge 2\;$$