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March 29, 2015

March 29, 2015

Posted by **Janice** on Monday, November 13, 2006 at 1:23am.

Riemann Sum [n=1 to infinity] 1/ [n(ln n) (ln(ln n))^p]

converge and for what values does it diverge?

You need to let the summation start at n = 3 to avoid the singularity at n = 1 (although you can formally take denominator to b infinity there...).

The series is divergent for all real p. do the integral test to see this: The series converges if and only if the integral

Int [x=3 to infinity] 1/[x(ln x) (ln(ln x))^p dx

converges. This is true if the terms in the series are positive and the function of x which you choose to match the terms of the series is a monotonic decreasing function, which is indeed the case here. We start at x = 3 because for n = 2 the term ln(ln(2)) is negative so for general the theorem isn't valid. But you only need to investigate a tail of the summation starting at some arbitrary n to infinity to extablish convergence or non-convergence.

If yom change varibles y = exp(exp(x) in the integral you see that the integral is:

Int [y=ln(ln(3) to infinity] y^(p)exp(y)dy

which is divergent for all p.

correction, the ln(ln(x)) factor is in the denominator and the integral is therefore:

Int [y=ln(ln(3) to infinity] y^(-p)exp(y)dy

which is still diverges for all p.

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