# Math Help - The Integral Test

1. ## The Integral Test

The question:

$\sum _{n=2}^{\infty }{\frac {1}{n \left( \ln \left( n \right)
\right) ^{2}}}$

I need help in finding if $f \left( x \right) ={\frac {1}{x \left( \ln \left( x \right)
\right) ^{2}}}
$
is decreasing.

I found that ${\frac {d}{dx}}f \left( x \right) =-{\frac { \left( \ln \left( x
\right) \right) ^{2}+2\,\ln \left( x \right) }{{x}^{2} \left( \ln
\left( x \right) \right) ^{4}}}$
. Then I made $-{\frac { \left( \ln \left( x \right) \right) ^{2}+2\,\ln \left( x
\right) }{{x}^{2} \left( \ln \left( x \right) \right) ^{4}}}<0
$

which turns out to be ${e}^{-2}

but when I graph it, f(x) is still increasing at ${e}^{-2}$. What am I doing wrong?

2. Originally Posted by s3n4te
The question:

$\sum _{n=2}^{\infty }{\frac {1}{n \left( \ln \left( n \right)
\right) ^{2}}}$

I need help in finding if $f \left( x \right) ={\frac {1}{x \left( \ln \left( x \right)
\right) ^{2}}}
$
is decreasing.

I found that ${\frac {d}{dx}}f \left( x \right) =-{\frac { \left( \ln \left( x
\right) \right) ^{2}+2\,\ln \left( x \right) }{{x}^{2} \left( \ln
\left( x \right) \right) ^{4}}}$
. Then I made $-{\frac { \left( \ln \left( x \right) \right) ^{2}+2\,\ln \left( x
\right) }{{x}^{2} \left( \ln \left( x \right) \right) ^{4}}}<0
$

which turns out to be ${e}^{-2}

but when I graph it, f(x) is still increasing at ${e}^{-2}$. What am I doing wrong?

But your sum starts at $n = 2$ so you only need to consider values of $x \ge 2$

3. Originally Posted by danny arrigo
But your sum starts at $n = 2$ so you only need to consider values of $x \ge 2$

I know, but I'm just puzzled why it's not decreasing at ${e}^{-2}$

4. Originally Posted by s3n4te

I know, but I'm just puzzled why it's not decreasing at ${e}^{-2}$

$y = \frac{1}{x(\ln{x})^2}$

$y' = \frac{-[2\ln{x} + (\ln{x})^2]}{x^2(\ln{x})^4}$

$y' = 0$ ...

$\ln{x}(2 + \ln{x}) = 0$

$\ln{x} = 0$ at $x = 1$ ... $y$ is undefined there

$\ln{x} = -2$

$x = e^{-2}$ ... $y$ is not decreasing at $x = e^{-2}$ since $y' = 0$ there.

5. Originally Posted by s3n4te

I know, but I'm just puzzled why it's not decreasing at ${e}^{-2}$
Your derivative (for $x \ne 1$)

$-{\frac { \left( \ln \left( x \right) \right) ^{2}+2\,\ln \left( x
\right) }{{x}^{2} \left( \ln \left( x \right) \right) ^{4}}} = -{\frac { \ln x +2}{{x}^{2} \left( \ln \left( x \right) \right) ^{3}}}$

and for $x < e^{-2}$ your denominator is negative make the entire expression positive.

6. Originally Posted by skeeter
$y = \frac{1}{x(\ln{x})^2}$

$y' = \frac{-[2\ln{x} + (\ln{x})^2]}{x^2(\ln{x})^4}$

$y' = 0$ ...

$\ln{x}(2 + \ln{x}) = 0$

$\ln{x} = 0$ at $x = 1$ ... $y$ is undefined there

$\ln{x} = -2$

$x = e^{-2}$ ... $y$ is not decreasing at $x = e^{-2}$ since $y' = 0$ there.
but if you make $y' < 0$...
${e}^{-2}

doesn't that mean right after ${e}^{-2}$ it should decrease?

7. Originally Posted by s3n4te
but if you make $y' < 0$...
${e}^{-2}

doesn't that mean right after ${e}^{-2}$ it should decrease?
between the two critical values, $x = e^{-2}$ and $x = 1$ is $x = e^{-1}$

... $y'(e^{-1}) > 0$ , which indicates y is increasing on that interval.

8. Originally Posted by skeeter
between the two critical values, $x = e^{-2}$ and $x = 1$ is $x = e^{-1}$

... $y'(e^{-1}) > 0$ , which indicates y is increasing on that interval.
Cool, but how am I supposed to figure that out with the equality test $-{\frac { \left( \ln \left( x \right) \right) ^{2}+2\,\ln \left( x
\right) }{{x}^{2} \left( \ln \left( x \right) \right) ^{4}}}<0
$
?

9. Originally Posted by s3n4te
Cool, but how am I supposed to figure that out with the equality test $-{\frac { \left( \ln \left( x \right) \right) ^{2}+2\,\ln \left( x
\right) }{{x}^{2} \left( \ln \left( x \right) \right) ^{4}}}<0
$
?
Never heard of the "equality" test. How do you do an "equality" test using an inequality?

Do you know the first derivative and second derivative tests for extrema?

10. Originally Posted by skeeter
Never heard of the "equality" test. How do you do an "equality" test using an inequality?

Do you know the first derivative and second derivative tests for extrema?
oops sorry I meant inequality :P

Yea but that was last term.... OH I get it now lol