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

Jan 28, 2017

${\sum}_{n = 1}^{\infty} \frac{1}{n \sqrt{n}}$ converges by the integral test.

#### Explanation:

At a glance, we can tell that the series converges if we use the handy fact that $\sum \frac{1}{n} ^ a$ converges if and only if $a > 1$. That the given series converges then is just a matter if noticing that $\frac{1}{n \sqrt{n}} = \frac{1}{n} ^ \left(\frac{3}{2}\right)$.

However, let's see how we could prove this using a test for convergence or divergence.

Using the integral test:

This test states that if $f \left(n\right) = {a}_{n}$ for all $n \in {\mathbb{Z}}^{+}$, then ${\sum}_{n = 1}^{\infty}$ converges if and only if ${\int}_{1}^{\infty} f \left(x\right) \mathrm{dx}$ converges. To apply this, we let $f \left(x\right) = \frac{1}{x \sqrt{x}} = {x}^{- \frac{3}{2}}$. Then

${\int}_{1}^{\infty} f \left(x\right) \mathrm{dx} = {\int}_{1}^{\infty} {x}^{- \frac{3}{2}} \mathrm{dx}$

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

$= - 2 {\left[\frac{1}{\sqrt{x}}\right]}_{1}^{\infty}$

$= - 2 \left(\frac{1}{\infty} - \frac{1}{1}\right)$

$= 2$

As ${\int}_{1}^{\infty} f \left(x\right) \mathrm{dx}$ converges and $f \left(n\right) = \frac{1}{n \sqrt{n}}$ at each positive integer $n$, ${\sum}_{n = 1}^{\infty} \frac{1}{n \sqrt{n}}$ converges by the integral test (note that it does not necessarily converge to the same value as the integral).

As a side note, we could substitute in $\frac{1}{x} ^ a$ for $\frac{1}{x} ^ \left(\frac{3}{2}\right)$, perform the same test, and find that the integral converges if and only if $a > 1$, verifying the shortcut mentioned in the start.