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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!

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

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  • $\begingroup$ thank you very much. i got out of my way and confused myself :'). iam sorry it might have seemed silly to ask but iam trying to follow my professor's method in identifying decreasing or increasing function using derivatives but this allowed me to think outside the box a bit. thank you again! ^^ $\endgroup$ – Aya Rahima Mar 29 at 16:13
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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 $\;x\ge 2\;$

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  • $\begingroup$ ooh iam sorry i just realized the derivative was wrong for missing the minus sign. thank you very much ! $\endgroup$ – Aya Rahima Mar 29 at 16:22

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