Find the interval of convergence, including end-point tests:$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{)}}^{\mathbf{n}}}{{\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{)}}^{\mathbf{n}}\sqrt{\mathbf{n}}}$

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

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

$$\begin{gathered} \rho =li{m}_{n\to \infty }\left({\rho }_n\right) \\ \rho =li{m}_{n\to \infty }\left(\frac{{\rho }_{n+1}}{{\rho }_n}\right) \end{gathered}$$.

Then the series$\sum _{n=0}^{\infty }{a}_n$ is convergent if $\rho <1$, divergent if$\rho >1$ and use anther set if $\rho =1$.

For end points test we use the following theorem:

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

The given series is:$\sum _{n=1}^{\infty }\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}$ .

Thus, we have$a_n=\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}$ and $a_{n+1}=\sum _{n=1}^{\infty }\frac{{(x+2)}^{n+1}}{{(-3)}^{n+1}\sqrt{n+1}}$.

Therefore, the ratio

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

Then,

$$\begin{gathered} \rho =\left({\rho }_n\right) \\ =li{m}_{n\to \infty }\left|\frac{(x+2)}{3}·\frac{\sqrt{n}}{\sqrt{n+1}}\right| \\ =li{m}_{n\to \infty }\left|\frac{(x+2)}{3}\right| \end{gathered}$$

The series is convergent when

$\rho <1$

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

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

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

$\begin{array}{rcl}\sum _{n=1}^{\infty }\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}& =& \sum _{n=1}^{\infty }\frac{{(-5+2)}^n}{{(-3)}^n\sqrt{n}}\\ & =& \sum _{n=1}^{\infty }\frac{{(-3)}^n}{{(-3)}^n\sqrt{n}}\\ & =& \sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}\\ & =& -\frac{1}{1}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\cdots \\ & & \end{array}$

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

Thus, the series $\sum _{n=1}^{\infty }\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\cdots \frac{{(-1)}^{n+1}}{2n-1}\cdots$diverges at lower endpoint $x=-5$.

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

$\begin{array}{rcl}\sum _{n=1}^{\infty }\frac{{(x+2)}^n}{{(-3)}^n\sqrt{n}}& =& \sum _{n=1}^{\infty }\frac{{(1+2)}^n}{{(-3)}^n\sqrt{n}}\\ & =& \sum _{n=1}^{\infty }\frac{{\left(3\right)}^n}{{(-3)}^n\sqrt{n}}\\ & =& \sum _{n=1}^{\infty }{(-1)}^n\frac{1}{\sqrt{n}}\\ & =& -\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\cdots \\ & & \end{array}$

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

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