# How do you test the series Sigma sqrt(n+1)-sqrtn from n is [0,oo) for convergence?

Feb 21, 2017

The series:

${\sum}_{n = 0}^{\infty} \sqrt{n + 1} - \sqrt{n}$

is divergent.

#### Explanation:

We can see that the series is divergent by analyzing the partial sums:

${s}_{n} = {\sum}_{k = 0}^{n} = \left(\cancel{1} - 0\right) + \left(\cancel{\sqrt{2}} - \cancel{1}\right) + \left(\sqrt{3} - \cancel{\sqrt{2}}\right) + \ldots + \left(\cancel{\sqrt{n}} - \sqrt{n - 1}\right) + \left(\sqrt{n + 1} - \cancel{\sqrt{n}}\right) = \sqrt{n + 1}$

so that:

${\lim}_{n \to \infty} {s}_{n} = {\lim}_{n \to \infty} \sqrt{n + 1} = \infty$

We can however use a convergence test in the following way: we have that:

$\sqrt{n + 1} - \sqrt{n} = \left(\sqrt{n + 1} - \sqrt{n}\right) \frac{\sqrt{n + 1} + \sqrt{n}}{\sqrt{n + 1} + \sqrt{n}}$

Using the identity:

$\left(a + b\right) \left(a - b\right) = \left({a}^{2} - {b}^{2}\right)$

this becomes:

$\sqrt{n + 1} - \sqrt{n} = \frac{\cancel{n} + 1 - \cancel{n}}{\sqrt{n + 1} + \sqrt{n}} = \frac{1}{\sqrt{n + 1} + \sqrt{n}}$

Now we can use the limit comparison test with the harmonic series:

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

which proves that:

${\sum}_{n = 0}^{\infty} \sqrt{n + 1} - \sqrt{n} = \infty$