Find the interval of convergence, including end-point tests$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{x}}^{\mathbf{n}}{\mathbf{n}}^{\mathbf{2}}}{{\mathbf{5}}^{\mathbf{n}}\left(n^2+1\right)}$

To find the interval of convergence of a given series, we use the 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}$$\rho =li{m}_{n\to \infty }\left({\rho }_n\right)=li{m}_{n\to \infty }\left(\frac{a_{n+1}}{a_n}\right)$and.

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

another set if $\rho =1$.

The given series is $\sum _{n=0}^{\infty }\frac{x^n{n}^2}{5^n\left(n^2+1\right)}$.

Thus, we have$a_n=\frac{x^n{n}^2}{5^n\left(n^2+1\right)}$ and $a_{n+1}=\frac{x^{n+1}{(n+1)}^2}{5^{n+1}\left({(n+1)}^2+1\right)}$.

Therefore, the ratio

$\begin{array}{rcl}{\rho }_n& =& \left|\frac{a_{n+1}}{a_n}\right|\\ & =& \left|\frac{\frac{x^{n+1}{(n+1)}^2}{5^{n+1}\left({(n+1)}^2+1\right)}}{\frac{x^n{n}^2}{\left.5n^2+1\right)}}\right|\\ & =& \left|\frac{x^{n+1}{(n+1)}^2}{x^n{n}^2}·\frac{5^n\left(n^2+1\right)}{5^{n+1}\left({(n+1)}^2+1\right)}\right|\\ & =& \left|\frac{x}{5}{\left(\frac{n+1}{n}\right)}^{2·}\left(\frac{n^2+1}{n^2+2n+1}\right)\right|\\ & & \end{array}$

Then,

role="math" localid="1664269502382" $$\begin{gathered} \rho =li{m}_{n\to \infty }\left({\rho }_n\right) \\ =li{m}_{n\to \infty }\left|\frac{x}{5}{\left(\frac{n+1}{n}\right)}^{2·}\left(\frac{n^2+1}{n^2+2n+1}\right)\right| \\ =li{m}_{n\to \infty }\left|\frac{x}{5}{\left(n+\frac{1}{n}\right)}^{2·}\left(\frac{1+\frac{1}{n^2}}{1+\frac{2}{n}+\frac{1}{n^2}}\right)\right| \\ =\left|\frac{x}{5}\right| \end{gathered}$$

The series is convergent when

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

At the endpoint

$\begin{array}{rcl}x& =& -5,\\ a_n& =& \frac{x^n{n}^2}{5^n\left(n^2+1\right)}\\ & =& \frac{{(-5)}^n{n}^2}{\left(n^2+1\right)}\\ & =& \frac{{(-1)}^n{n}^2}{\left(n^2+1\right)}\\ & & \end{array}$.

$\begin{array}{rcl}\sum _{n=0}^{\infty }\frac{{(-1)}^n{n}^2}{\left(n^2+1\right)}& =& \sum _{n=0}^{\infty }\frac{{(-1)}^n{n}^2+1-1}{\left(n^2+1\right)}\\ & =& \sum _{n=0}^{\infty }\left\{{(-1)}^n-\frac{{(-1)}^n}{\left(n^2+1\right)}\right\}\\ & =& \sum _{n=0}^{\infty }{(-1)}^n+\sum _{n=0}^{\infty }-\frac{{(-1)}^n}{\left(n^2+1\right)}\\ & =& \sum _{n=0}^{\infty }{A}_n+\sum _{n=0}^{\infty }{B}_n\\ & & \end{array}$

Since $\sum _{n=0}^{\infty }{A}_n$is divergent so the series $\sum _{n=0}^{\infty }\frac{{(-1)}^n{n}^2}{\left(n^2+1\right)}$, thus, at endpoint$x=-5$ the series is divergent.

At the endpoint $x=5$, the series$\sum _{n=0}^{\infty }\frac{n^2}{\left(n^2+1\right)}==\sum _{n=0}^{\infty }\left\{1-\frac{1}{\left(n^2+1\right)}\right\}$is divergent as

$\sum _{n=0}^{\infty }1=[n{]}^{\infty }=\infty$.

Moreover, when $n\to \infty ,\frac{n^2}{n^2+1}~1$.

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