The Integral test sum[n=2,infty][1/(n (ln n)^2)]

[maeveoneill]

Determine whether the series is convergent or divergent.

infinity
-----
\..........1/[n (ln n)^2]
/
---
n=2

Solution.
we must prove the function is continuous, positive and decreasing to use the integral test..

so we know it is positve and continuous by looking at it. to prove it is decreasing we find the derivative.

this is what i have so far when trying to find the deriviateve::

f'(x) = [x(ln x)^2 (0) - [ (lnx)^2 (1) + 2xlnx (1/x)]]/ [x(lnx)^2]^2

= [-(lnx)^2 - [(2xlnx)/x]]/ [x(lnx)^2]^2

= [-(lnx)^2 - 2lnx] / [x(lnx)^2]^2

is this right so far, and then can i divide everything by ln??

 

[stapel]

...= [-(lnx)^2 - 2lnx] / [x(lnx)^2]^2

is this right so far, and then can i divide everything by ln??

That's what I get. Then you divide out the common factor of ln(x) to get:

. . . . .\(\displaystyle \L f'(x)\, =\, \frac{-ln(x)\, -\, 2}{x^2\,ln^3(x)}\)

You need to show that this is always negative, so that f(x) is always decreasing. Whatever x is, x² is positive. Since you only care x > 2 (since n is specified to be such), then you know that ln(x) must be positive, and so is its cube. So the denominator is positive.

What can you say about the numerator, and thus the entire fraction? :wink:

Eliz.

 

[maeveoneill]

The numerator will always be negative. Thus a negative divided by a postive will always be negative.. so the function is decreasing?

I have a solution book and it says

f'(x) = - [p + lnx]/ [(x^2 )(lnx)^(p+1)] <0 if x > e^-p

where deoes this e come from?? shouldnt it be less tahn 0 for all x>/2

 

[Deleted member 4993]

The numerator will always be negative. Thus a negative divided by a postive will always be negative.. so the function is decreasing?

I have a solution book and it says

f'(x) = - [p + lnx]/ [(x^2 )(lnx)^(p+1)] <0 if x > e^-p implies ln(x) + p > 0

where deoes this e come from?? shouldnt it be less tahn 0 for all x>/2

 

[pka]

\(\displaystyle \L \int\limits_2^b {\frac{{dx}}{{x\ln ^2 (x)}}} = \left. {\frac{{ - 1}}{{\ln (x)}}} \right|_{x = 2}^{x = b}\)