# How do you use the limit comparison test on the series sum_(n=2)^oosqrt(n)/(n-1) ?

Jul 29, 2015

The series diverges.

#### Explanation:

Note first that we can write:

${\sum}_{n = 2}^{\infty} \frac{\sqrt{n}}{n - 1}$ as ${\sum}_{n = 1}^{\infty} \frac{\sqrt{n + 1}}{n}$

In the limit, the terms 'act like' $\frac{1}{\sqrt{n}}$ so we'll use the series:

${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n}}$ for the limit comparison test.

$c = {\lim}_{n \rightarrow \infty} \frac{\frac{\sqrt{n + 1}}{n}}{\frac{1}{\sqrt{n}}}$

$= {\lim}_{n \rightarrow \infty} \frac{\sqrt{n + 1} \sqrt{n}}{n}$

$= {\lim}_{n \rightarrow \infty} \frac{\sqrt{{n}^{2} + n}}{n}$

$= {\lim}_{n \rightarrow \infty} \frac{n \sqrt{1 + \left(\frac{1}{n}\right)}}{n}$

$= 1$

Since $c$ is positive and finite the two series either both converge or both diverge.

${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n}}$ diverges by comparison with the harmonic series, so

${\sum}_{n = 1}^{\infty} \frac{\sqrt{n + 1}}{n}$ diverges, as does the original series.

${\sum}_{n = 2}^{\infty} \frac{\sqrt{n}}{n - 1}$ diverges.