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

Feb 2, 2017

The series:

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

is convergent

#### Explanation:

The series:

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

is an alternating series, so we can test its convergence using Leibniz's theorem, which states that an alternating series

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {a}_{n}$

is convergent if:

(i) ${\lim}_{n \to \infty} {a}_{n} = 0$

(ii) ${a}_{n + 1} \le {a}_{n}$

in our case:

${\lim}_{n \to \infty} \frac{\sqrt{n}}{n + 1} = {\lim}_{n \to \infty} \frac{1}{\frac{n + 1}{\sqrt{n}}} = {\lim}_{n \to \infty} \frac{1}{\left(\sqrt{n} + \frac{1}{\sqrt{n}}\right)} = 0$

so the first condition is satisfied.
For the second we analyze the function:

$f \left(x\right) = \frac{\sqrt{x}}{x + 1}$

and calculate the derivative:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{x + 1}{2 \sqrt{x}} - \sqrt{x}}{x + 1} ^ 2 = \frac{x + 1 - 2 x}{2 \sqrt{x} {\left(x + 1\right)}^{2}} = - \frac{x - 1}{2 \sqrt{x} {\left(x + 1\right)}^{2}}$

we can see that $\frac{\mathrm{df}}{\mathrm{dx}} < 0$ for $x \in \left(1 , + \infty\right)$ therefore the function is strictly decreasing in that interval and we have:

$f \left(n + 1\right) < f \left(n\right)$

that is:

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

and also the second condition is satisfied.