What values of x makes the series converge (ln(n)^x)/n

Question

Series starts from 2 to infinity

Answer 1

Based on the requested series --> f(n) = ln^x(n)/n we want to understand the "strength of the functions" involved (assumption is that n > 0 since the logarithm is not defined).

The general order of strength of types of functions follows as such...

Exponentials

Polynomials

Sinusoidals

Logarithms

In this case we have two functions : Logarithm and Polynomial

Let's look at some simple case(s) to see if there is clear breakout...

First, if x = 0 then the function looks like the harmonic function ( 1/n ). This series is known by the p-test to not converge (divergent). If x < 0 (i.e. negative), the logarithm moves into the denominator and will cause the series to converge since it is slightly "stronger" than just n ( i.e. 1/(n*ln^x(n)) ). We also know the logarithm if a strictly increasing function (from n >0⁺).

Therefore, half the available x values have been determined in terms of convergence and divergence. Let's now look into the positive half of the x-axis...

Due to the fact that the logarithm is a strictly increasing function we can look at the ratio-test of series to determine what values of x will allow for absolute convergence and conditional convergence...

lim |a_n+1/a_n| = | ln^x(n+1)/(n+1) / ln^x(n)/n | = | n/n+1 * ln^x(n+1)/ln^x(n) |

n --> infinity

The limit of n/(n+1) as n approached infinity is 1. What about the ratio of the logarithms?

ln^x(n+1)/ln^x(n) = (ln(n+1)/ln(n))^x

This limit is also 1 due to the ratio of logarithms

So we come to conclusion that the limit = 1 which is inconclusive. So let's look at the integral test...

Let's define x --> a ( a parameter )

Let's define n --> x ( the variable to integrate over )

a > 0

integral [lb=0, ub=infinity] (ln^a(x)/x * dx)

Integrating yields --> ln^(a+1)(x)/a

We can see that for any value of "a" will cause this integral to continue to expand without bound. Therefore, the only values which allow allow for convergence are x < 0.