For what values of p>0 does the series
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.
To determine whether the series
Riemann Sum [n=3 to infinity] 1/ [n(ln n) (ln(ln n))^p]
converges or diverges, we can use the integral test.
The integral test states that if a series is positive and the terms of the series match the terms of a function that is continuous, positive, and decreasing on the interval [3, infinity), then the series and the integral of the function will both converge or both diverge.
Let's consider the integral of the function f(x) = 1/[x(ln x)(ln(ln x))^p] from x = 3 to infinity:
∫[x=3 to infinity] 1/[x(ln x)(ln(ln x))^p] dx
To simplify the integral, we can change variables by letting y = exp(exp(x)). Then, dx = dy/(y(ln y)).
The integral becomes:
∫[y=ln(ln(3)) to infinity] (y^(-p)exp(y))/y dy
Now, we need to determine whether this integral converges or diverges for different values of p.
If we analyze the behavior of the integrand as y approaches infinity, we see that y^(-p)exp(y) dominates the denominator. Since y^(-p) approaches infinity as y approaches infinity, the integral diverges for all positive values of p. Therefore, the series
Riemann Sum [n=3 to infinity] 1/ [n(ln n) (ln(ln n))^p]
also diverges for all positive values of p.