Question

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathbf{x}}^{\mathbf{2}\mathbf{n}}}{{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}}^{{\displaystyle \frac{3}{2}}}}$

Short answer

The interval of convergence is $\left[-1,1\right]$

Step-by-step solution

Given Information   

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

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

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

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

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

$=\frac{{\displaystyle \frac{x^{2n+2}}{{\left(2n+2\right)}^{{\displaystyle \frac{3}{2}}}}}}{{\displaystyle \frac{x^{2n}}{{\left(2n\right)}^{{\displaystyle \frac{3}{2}}}}}}$

$=\left|\frac{x^{2n+2}{\left(2n\right)}^{{\displaystyle \frac{3}{2}}}}{x^{2n}{\left(2n+2\right)}^{{\displaystyle \frac{1}{2}}}}\right|$

$=\left|\frac{x^2{2}^{{\displaystyle \frac{3}{2}}}}{{\left(2+{\displaystyle \frac{2}{n}}\right)}^{{\displaystyle \frac{3}{2}}}}\right|$

Apply limits in the above equation,

$\rho =\underset{n\to \infty }{\lim }\left|\frac{x^2{2}^{{\displaystyle \frac{3}{2}}}}{{\left(2+{\displaystyle \frac{2}{n}}\right)}^{{\displaystyle \frac{3}{2}}}}\right|$

$=\left|x^2\right|$

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

Hence the given power series is convergent for $\left|x^2\right|<1$and divergent for $\left|x^2\right|>1$.

Now check the ends points,

When $x=1$series is $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{n^{{\displaystyle \frac{3}{2}}}}$, which is a convergent alternating series.

When $x=-1$series is $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{n^{{\displaystyle \frac{3}{2}}}}$, which is a convergent alternating series.

Hence the interval of convergence is $\left[-1,1\right]$.