# How do you find the positive values of p for which Sigma lnn/n^p from [2,oo) converges?

Feb 3, 2017

The series:

${\sum}_{n = 2}^{\infty} \ln \frac{n}{n} ^ p$

is convergent for $p > 1$.

#### Explanation:

A necessary condition for a series to converge is that its general term is infinitesimal, that is:

(1) ${\lim}_{n \to \infty} \ln \frac{n}{n} ^ p = 0$

Consider the function $f \left(x\right) = \ln \frac{x}{x} ^ p$ and the limit:

(2) ${\lim}_{x \to \infty} \ln \frac{x}{x} ^ p$

Clearly if such limit exists it must coincide with (1).

Now, for $p = 0$ the limit (2) is $\infty$, while for $p \ne 0$ it is in the indeterminate form $\frac{\infty}{\infty}$ so we can solve it using l'Hospital's rule:

${\lim}_{x \to \infty} \ln \frac{x}{x} ^ p = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} {x}^{p}} = {\lim}_{x \to \infty} \left(\frac{1}{x}\right) \left(\frac{1}{p {x}^{p - 1}}\right) = {\lim}_{x \to \infty} \frac{1}{p {x}^{p}} = \left\{\begin{matrix}0 \text{ for " p > 0 \\ oo " for } p < 0\end{matrix}\right.$

So we know that for $p \le 0$ the series is not convergent.

To have a sufficient condition we can apply the integral test, using as test function: $f \left(x\right) = \ln \frac{x}{x} ^ p$.

For $p > 0$ such function is positive, decreasing and, as we have just seen, infinitesimal, so it satisfies the hypotheses of the integral test theorem, and the series is proven to be convergent if the improper integral:

${\int}_{2}^{\infty} \ln \frac{x}{x} ^ p$

also converges.

Solving the indefinite integral by parts:

$\int \ln \frac{x}{x} ^ p \mathrm{dx} = \int \ln x d \left({x}^{1 - p} / \left(1 - p\right)\right) = \ln x {x}^{1 - p} / \left(1 - p\right) - \int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx} = \frac{1}{1 - p} \ln \frac{x}{x} ^ \left(p - 1\right) - \int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx}$

$\int \frac{1}{x} {x}^{1 - p} / \left(1 - p\right) \mathrm{dx} = \frac{1}{p - 1} \int {x}^{- p} \mathrm{dx} = \frac{1}{1 - p} ^ 2 {x}^{1 - p} + C = \frac{1}{1 - p} ^ 2 \frac{1}{x} ^ \left(p - 1\right) + C$

So that:

${\int}_{2}^{\infty} \ln \frac{x}{x} ^ p = {\left[\frac{1}{1 - p} \ln \frac{x}{x} ^ \left(p - 1\right) - \frac{1}{1 - p} ^ 2 \frac{1}{x} ^ \left(p - 1\right)\right]}_{2}^{\infty}$

which is convergent for $p - 1 > 0 \implies p > 1$

In conclusion the series is convergent for $p > 1$