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

Nov 1, 2017

See below.

#### Explanation:

We have

$\sqrt{n + 1} - \sqrt{n} \le \frac{1}{2} \frac{1}{\sqrt{n}}$ so the series

${\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{n} \left(\sqrt{n + 1} - \sqrt{n}\right) \le \frac{1}{2} {\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{n} \frac{1}{\sqrt{n}}$

but the series with ${a}_{n} = {\left(- 1\right)}^{n} \frac{1}{\sqrt{n}}$ is alternating and absolutely convergent then

${\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{n} \left(\sqrt{n + 1} - \sqrt{n}\right)$ is convergent.

NOTE:

From the mean value theorem, with $f \left(x\right) \in {\mathbb{C}}^{2}$ there exists $\xi \in \left(a , b\right)$ such that

$f \left(b\right) - f \left(a\right) = f ' \left(\xi\right) \left(b - a\right)$

Calling $f \left(x\right) = \sqrt{x}$, $a = n$, $b = n + 1$ and knowing that in this case

$f ' \left(x\right)$ is monotonically decreasing in $n , n + 1$ we establish that

$\sqrt{n + 1} - \sqrt{n} \le \frac{1}{2} \frac{1}{\sqrt{n}}$