$\sum _{n=1}^{\mathrm{\infty }} {\left(\sqrt{x^2+1}\right)}^n\frac{2^n}{3^n+n^3}$

The power series is$\sum _{n=1}^{\mathrm{\infty }} {\left(\sqrt{x^2+1}\right)}^n\frac{2^n}{3^n+n^3}$

The Interval of convergence is the interval in which the power series is convergent.

The power series is $\sum _{n=1}^{\mathrm{\infty }} {\left(\sqrt{x^2+1}\right)}^n\frac{2^n}{3^n+n^3}$

Let,$y=\sqrt{x^2+1}$

The power series becomes $\sum _{n=1}^{\mathrm{\infty }} y^n\frac{2^n}{3^n+n^3}$

$\text{Let}{\rho }_n=\left|\frac{a_{n+1}}{a_n}\right|\text{,}$

Substitute the value of the power series in the formula above, the equation becomes as follows,

$\begin{array}{r}{\rho }_n=\left|\frac{a_{n+1}}{a_n}\right|\\ =\left|\frac{2^{n+1}{y}^{n+1}}{\frac{2^{n+1}{y}^n}{3^n+n^3}}\right|\\ =\left|\frac{2^{n+1}{y}^{n+1}\left[3^n+n^3\mid \right.}{\left.3^{n+1}+(n+1{)}^3\right]2^n{y}^n}\right|\\ =\left|\frac{2y\left[3^n+n^3\right]}{\left(3^{n+1}+(n+1{)}^3\right]}\right|\end{array}$

Apply limits in the above equation

$\begin{array}{r}\rho =\left|2y\right|\underset{n\to \mathrm{\infty }}{lim} \frac{3^n+n^3}{3^{n+1}+(n+1{)}^3}\\ =\left|2y\right|\underset{n\to \mathrm{\infty }}{lim} \frac{(\ln (3\left)\right)3^n+3n^2}{(\ln (3\left)\right)3^{n+1}+3(n+1{)}^2}\\ =\left|2y\right|\underset{n\to \mathrm{\infty }}{lim} \frac{(\ln (3){)}^3{3}^n+6}{(\ln (3){)}^3{3}^{n+1}+6}\\ =\left|\frac{2y}{3}\right|\end{array}$

The power series is convergent for $\rho <1$

Hence the given power series is convergent for$\left|\frac{2y}{3}\right|<1$and divergent for$\left|\frac{2y}{3}\right|>1$

The series is convergent when$\left|\frac{2y}{3}\right|<1$

$\begin{array}{c}\left|\frac{2y}{3}\right|<1\\ -3<2y<3\mid \\ \frac{-3}{2}<y<\frac{3}{2}\end{array}$

Find the limits of x,

$\begin{array}{r}-\frac{3}{2}<\sqrt{x^2+1}<\frac{3}{2}\\ \frac{-9}{4}<x^2+1<\frac{9}{4}\\ \frac{-\sqrt{5}}{2}<x<\frac{\sqrt{5}}{2}\end{array}$

Now check the ends points,

When $x=\frac{-\sqrt{5}}{2}\text{series is}\sum _{n=1}^{\mathrm{\infty }} \frac{3^n}{3^n+n^3}$, which is a divergent series.

When X$=\frac{\sqrt{5}}{2}\text{series is}\sum _{n=1}^{\mathrm{\infty }} \frac{3^n}{3^n+n^3}$ , which is a divergent series.

Hence the interval of convergence is $\left(-\frac{\sqrt{5}}{2},\frac{\sqrt{5}}{2}\right)$