91Ó°ÊÓ

To find the interval of convergence of a given series, we use ratio test stated below:

Let ${\mathrm{ÒÏ}}_n$be the ratio of two successive terms of an infinite series ${\mathrm{ÒÏ}}_n=\frac{a_{n+1}}{a_n}$and

$$\begin{gathered} \mathrm{ÒÏ}=li{m}_{n→∞}\left({\mathrm{ÒÏ}}_n\right) \\ \mathrm{ÒÏ}=li{m}_{n→∞}\left(\frac{{\mathrm{ÒÏ}}_{n+1}}{{\mathrm{ÒÏ}}_n}\right) \end{gathered}$$.

Then the series$\underset{n=0}{\overset{∞}{∑}}{a}_n$ is convergent if $\mathrm{ÒÏ}<1$, divergent if$\mathrm{ÒÏ}>1$ and use anther set if $\mathrm{ÒÏ}=1$.

For end points test we use the following theorem:

An infinite series $\underset{\mathbf{n}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathit{b}}_{\mathbf{n}}$in which the terms are alternately positive and negative is convergent if each term is numerically less than the preceding term and$li{m}_{n→∞}{a}_n=0$.

The given series is:$\underset{n=1}{\overset{∞}{∑}}\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}$ .

Thus, we have$a_n=\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}$ and $a_{n+1}=\underset{n=1}{\overset{∞}{∑}}\frac{{(x+2)}^{n+1}}{{(-3)}^{n+1}\sqrt{n+1}}$.

Therefore, the ratio

$\begin{array}{rcl}{\mathrm{ÒÏ}}_n& =& \left|\frac{a_{n+1}}{a_n}\right|\\ & =& \left|\frac{\frac{{(x+2)}^{n+1}}{{(-3)}^{n+1}\sqrt{n+1}}}{\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}}\right|\\ & =& \left|\frac{{(x+2)}^{n+1}}{{(-3)}^{n+1}\sqrt{n+1}}·\frac{{(-3)}^n\sqrt{n}}{{(x+2)}^n}\right|\\ & =& \left|\frac{(x+2)}{-3}·\frac{\sqrt{n}}{\sqrt{n+1}}\right|\\ & =& \left|\frac{(x+2)}{3}·\frac{\sqrt{n}}{\sqrt{n+1}}\right|\\ & & \end{array}$

Then,

$$\begin{gathered} \mathrm{ÒÏ}=\left({\mathrm{ÒÏ}}_n\right) \\ =li{m}_{n→∞}\left|\frac{(x+2)}{3}·\frac{\sqrt{n}}{\sqrt{n+1}}\right| \\ =li{m}_{n→∞}\left|\frac{(x+2)}{3}\right| \end{gathered}$$

The series is convergent when

$\mathrm{ÒÏ}<1$

$$\begin{gathered} ⇒\left|\frac{(x+2)}{3}\right|<1 \\ ⇒-1<\frac{(x+2)}{3}<1 \\ ⇒-5<x<1. \end{gathered}$$

Thus, the given series is convergent in the interval $-5<x<1$.

At the end point $x=-5$, the series is

$\begin{array}{rcl}\underset{n=1}{\overset{∞}{∑}}\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}& =& \underset{n=1}{\overset{∞}{∑}}\frac{{(-5+2)}^n}{{(-3)}^n\sqrt{n}}\\ & =& \underset{n=1}{\overset{∞}{∑}}\frac{{(-3)}^n}{{(-3)}^n\sqrt{n}}\\ & =& \underset{n=1}{\overset{∞}{∑}}\frac{1}{\sqrt{n}}\\ & =& -\frac{1}{1}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+⋯\\ & & \end{array}$

As the series $S=\frac{1}{1^p}+\frac{1}{2^p}+\frac{1}{3^p}+⋯$diverges when $p<1$.

Thus, the series $\underset{n=1}{\overset{∞}{∑}}\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-⋯\frac{{(-1)}^{n+1}}{2n-1}⋯$diverges at lower endpoint $x=-5$.

At the end point $x=1$, the series is.

$\begin{array}{rcl}\underset{n=1}{\overset{∞}{∑}}\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}& =& \underset{n=1}{\overset{∞}{∑}}\frac{{(1+2)}^n}{{(-3)}^n\sqrt{n}}\\ & =& \underset{n=1}{\overset{∞}{∑}}\frac{{\left(3\right)}^n}{{(-3)}^n\sqrt{n}}\\ & =& \underset{n=1}{\overset{∞}{∑}}{(-1)}^n\frac{1}{\sqrt{n}}\\ & =& -\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+⋯\\ & & \end{array}$

It has alternately negative and positive terms, each term is less than the preceding term is and $li{m}_{n→∞}\left(\frac{1}{\sqrt{n}}\right)=0$, therefore, this series at the upper end-point$x=1$ is a converging series.

Hence, the given series is convergent in interval $-5<xâ\copyright ½1$.