# How do you test the series Sigma lnn/(nsqrtn) from n is [1,oo) for convergence?

Feb 3, 2017

The series:

${\sum}_{n = 1}^{\infty} \ln \frac{n}{n \sqrt{n}}$

is convergent.

#### Explanation:

You can use the integral test, choosing as test function:

$f \left(x\right) = \ln \frac{x}{x \sqrt{x}}$

We can verify that:

(i) $f \left(x\right) > 0$ for $x \in \left(1 , + \infty\right)$

(ii) ${\lim}_{x \to \infty} \ln \frac{x}{x \sqrt{x}} = 0$

The limit is in the indeterminate form $\frac{\infty}{\infty}$ so we can use l'Hospital's rule to solve it:

${\lim}_{x \to \infty} \ln \frac{x}{x \sqrt{x}} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} {x}^{\frac{3}{2}}} = {\lim}_{x \to \infty} \left(\frac{1}{x}\right) \frac{1}{\frac{3}{2} {x}^{\frac{1}{2}}} = {\lim}_{x \to \infty} \frac{2}{3 {x}^{\frac{3}{2}}} = 0$

(iii) $f \left(x\right)$ is monotone decreasing for $x \in \left(1 + \infty\right)$

We can calculate the first derivative:

$\frac{d}{\mathrm{dx}} \ln \frac{x}{x \sqrt{x}} = \frac{d}{\mathrm{dx}} \ln x \cdot {x}^{- \frac{3}{2}} = \left(\frac{1}{x}\right) {x}^{- \frac{3}{2}} - \frac{3}{2} {x}^{- \frac{5}{2}} \ln x = {x}^{- \frac{5}{2}} \left(1 - \frac{3}{2} \ln x\right) = \frac{2 - 3 \ln x}{2 {x}^{\frac{5}{2}}}$

and we can see that $x > 1 \implies f ' \left(x\right) < 0$

(iv) $f \left(n\right) = \ln \frac{n}{n \sqrt{n}}$

Thus all the hypotheses of the integral test theorem are satisfied and the convergence of the series is equivalent to the convergence of the improper integral:

${\int}_{1}^{\infty} \ln \frac{x}{x \sqrt{x}} \mathrm{dx}$

We solve the indefinite integral by parts:

$\int \ln x \cdot {x}^{- \frac{3}{2}} \mathrm{dx} = \int \ln x d \left(- 2 {x}^{- \frac{1}{2}}\right) = - 2 \ln \frac{x}{\sqrt{x}} + 2 \int {x}^{- \frac{1}{2}} / x \mathrm{dx} = - 2 \ln \frac{x}{\sqrt{x}} + 2 \int {x}^{- \frac{3}{2}} \mathrm{dx} = - 2 \ln \frac{x}{\sqrt{x}} - 4 {x}^{- \frac{1}{2}} + C = - \frac{2 \left(\ln x + 2\right)}{\sqrt{x}} + C$

and we have:

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

${\int}_{1}^{\infty} \ln \frac{x}{x \sqrt{x}} \mathrm{dx} = 4 - {\lim}_{x \to \infty} \ln \frac{x}{x \sqrt{x}}$

We have already calculated at point (ii) above that such limit is zero, then:

${\int}_{1}^{\infty} \ln \frac{x}{x \sqrt{x}} \mathrm{dx} = 4$

proving that the series is convergent.