Question

Test the following series for convergence

$\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{5}}$

Short answer

The alternating series$\sum _{\mathrm{n}-1}^{\infty }\frac{{(-1)}^{\mathrm{n}}\mathrm{n}}{\mathrm{n}+5}$ diverges.

Step-by-step solution

Significance of alternating series

If the terms' absolute values gradually decline to zero, or if$\mathbf{\left|}{\mathbf{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{\le }\mathbf{\left|}{\mathbf{a}}_{\mathbf{n}}\mathbf{\right|}$and$\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$, then an alternating series converges.

Use a test to check for convergence

Consider the alternating series,

$\sum _{\mathrm{n}-1}^{\infty }\frac{{(-1)}^{\mathrm{n}}\mathrm{n}}{\mathrm{n}+5}$

Use the below test to check the series for convergence:

$\left|a_{n+1}\right|\mathbf{\le }\left|a_n\right|$and$\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathbf{.}$

The series converges if the condition is met.

$$\begin{gathered} a_n=\frac{n{(-1)}^n}{n+5} \\ \left|a_n\right|=\frac{n}{n+5} \\ \left|a_{n+1}\right|=\frac{n+1}{n+6} \end{gathered}$$

Perform the test to check if both conditions are met

Let check if$\left|a_n\right|>\left|a_{n+1}\right|$ and using the following method.

For$\left|a_{n+1}\right|-\left|a_n\right|<0$ this will mean that$\left|a_n\right|>\left|a_{n+1}\right|$

For$\left|a_{n+1}\right|-\left|a_n\right|>0$ this will mean that$\left|a_n\right|<\left|a_{n+1}\right|$

$$\begin{gathered} \left|a_{n+1}\right|-\left|a_n\right|=\frac{n+1}{n+6}-\frac{n}{n+5} \\ =\frac{n^2+6n+5-n^2-6n}{(n+5)(n+6)} \\ =\frac{5}{(n+5)(n+6)} \end{gathered}$$

Now, for$n\ge 1,\left|a_{n+1}\right|-\left|a_n\right|>0,\left|a_n\right|<\left|a_{n+1}\right|$

Therefore, we conclude that$\left|a_{n+1}\right|>\left|a_n\right|$ for$n\ge 1$

As the first condition is not met, so the series diverges.