91Ó°ÊÓ

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

Let consider the alternating series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n\sqrt{10n}}{n+2}$

The series needs to be tested for convergence by using the following test:

$\mathbf{\left|}{\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{≤}\mathbf{|}{\mathit{a}}_{\mathbf{n}}$$\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{\left|}{\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{≤}\mathbf{|}{\mathit{a}}_{\mathbf{n}}$

The series converges upon satisfying the conditions.

$$\begin{gathered} a_n=\frac{{\left(-1\right)}^n\sqrt{10n}}{n+2} \\ \left|a_n\right|=\frac{\sqrt{10n}}{n+2} \\ \left|a_{n+1}\right|=\frac{\sqrt{10\left(n+1\right)}}{n+3} \end{gathered}$$

Check if$\left|a_n\right|>\left|a_{n+1}\right|$ using the following method:

1. If $\frac{|a_n+1|}{\left|a_n\right|}$, then $\left|a_n\right|>\left|a_{n+1}\right|$

2. If$\frac{|a_n+1|}{\left|a_n\right|}$ , then$\left|a_n\right|>\left|a_{n+1}\right|.$

Use the above method for the series get:

$$\begin{gathered} \frac{|a_n+1|}{\left|a_n\right|}=\frac{{\displaystyle \frac{\sqrt{10\left(n+1\right)}}{n+3}}}{{\displaystyle \frac{\sqrt{10n}}{n+2}}} \\ \frac{|a_n+1|}{\left|a_n\right|}=\frac{\left(n+2\right)\sqrt{\left(n+1\right)}}{\left(n+3\right)\sqrt{n}} \end{gathered}$$

From the expression in the previous step, Conclude that$(n+3)\sqrt{n}>(n+2)\sqrt{(n+1)}.$ . Thus,

role="math" localid="1658729220434" $\frac{\left(n+2\right)\sqrt{\left(n+1\right)}}{\left(n+3\right)\sqrt{n}}<1,\left|a_{n+1}\right|<\left|a_n\right|$

role="math" localid="1658729308511" $\underset{\mathrm{n}→∞}{lim}{a}_{\mathrm{n}}=\underset{\mathrm{n}→∞}{\lim }\frac{\sqrt{10n}}{n+2}$

This is an indeterminate kind of limit, so divide both the denominator and the numerator by to get the answer:

$$\begin{gathered} \underset{\mathrm{n}→∞}{lim}{a}_{\mathrm{n}}=\underset{\mathrm{n}→∞}{\lim }\frac{\sqrt{{\displaystyle \frac{10}{n}}}}{1+{\displaystyle \frac{2}{n}}} \\ =\frac{0}{1} \\ =0 \end{gathered}$$

All the conditions are met, so the series converges.