# For what values of p>0 does the series

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

converge and for what values does it diverge?

## To determine the convergence or divergence of the series

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

we can use the integral test. The integral test states that if f(x) is a positive, continuous, and decreasing function for x >= 1, then the series

Sum [n=1 to infinity] f(n)

converges if and only if the improper integral

Integral [1 to infinity] f(x) dx

converges.

Applying the integral test to the given series, we need to examine the convergence of the improper integral

Integral [1 to infinity] (1/[x(ln x)^p]) dx.

We can integrate this improper integral using the substitution u = ln x. This leads to the integral becoming

Integral [0 to infinity] (1/[u^p]) du.

Let's break down the possible values of p:

1. For p > 1:
As p > 1, the integral becomes
Integral [0 to infinity] (1/[u^p]) du = [u^-(p-1)] / (p - 1) evaluated from 0 to infinity.

Taking the limits of this integral, we find that it converges if p > 1.

2. For p <= 1:
As p <= 1, the integral becomes
Integral [0 to infinity] (1/[u^p]) du = [u^-(p-1)] / (p - 1) evaluated from 0 to infinity.

Taking the limits of this integral, we find that it diverges if p <= 1.

Therefore, the series Riemann Sum [n=1 to infinity] 1/ [n(ln n)^p] converges if p > 1 and diverges if p <= 1.