Question

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathit{n}}^{\mathbf{3}}{\mathit{x}}^{\mathbf{n}}$

Short answer

The interval of convergence is$\left(-1,1\right)$

Step-by-step solution

Given Information  

The power series is$\sum _{n=1}^{\infty }{\left(-1\right)}^n{n}^3{x}^n$

Definition of the interval of convergence. 

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

Find the interval. 

The power series is $\sum _{n=1}^{\infty }{\left(-1\right)}^n{n}^3{x}^n$

Let $p_n=\left|\frac{a_{n-1}}{a_n}\right|$.

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

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

$=\left|\frac{(n+1{)}^3{x}^{n+1}}{n^3{x}^n}\right|$

$=\left|\frac{(n+1{)}^{3}x}{n^3}\right|$

Apply limits in the above equation,

$\rho \underset{n\to \infty }{\lim }=\left|{\left(1+\frac{1}{n}\right)}^{3}x\right|$

$=\left|x\right|$

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

Hence the given power series is convergent for $\left|x\right|<1$and divergent for $\left|x\right|>1$.

Now check the ends points,

When $x=1$series is $\sum _{n=1}^{\infty }{\left(-1\right)}^n{n}^3$, which is a divergent alternating series.

When $x=-1$series is$\sum _{n=1}^{\infty }{n}^3$, which is a divergent series.

Hence the interval of convergence is $\left(-1,1\right)$.