# How do you determine if the series the converges conditionally, absolutely or diverges given Sigma ((-1)^(n))/(sqrt(n+4)) from [1,oo)?

Aug 8, 2017

The series:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} / \sqrt{n + 4}$

is convergent but not absolutely convergent.

#### Explanation:

This is an alternating series in the form:

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

thus based on Leibniz's theorem, it is convergent if:

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

$\left(2\right) {a}_{n} \ge {a}_{n + 1}$

In our case:

${\lim}_{n \to \infty} \frac{1}{\sqrt{n + 4}} = 0$

and

$\frac{1}{\sqrt{n + 4}} > \frac{1}{\sqrt{n + 5}}$

so both conditions are satisfied and the series is convergent.

Consider now the series of absolute values:

$\left(3\right) {\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n + 4}}$

Using the limit comparison test we can see that as:

${\lim}_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n + 4}} = 1$

the series $\left(3\right)$ has the same character as the series:

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

which is divergent based on the $p$-series test.

The series $\left(3\right)$ is then also divergent, so the series:

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} / \sqrt{n + 4}$

is convergent but not absolutely convergent.