Question

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{x}}^{\mathbf{2}\mathbf{n}}}{{\mathbf{2}}^{\mathbf{n}}{\mathbf{n}}^{\mathbf{2}}}$

Short answer

The convergence is between the intervals$\left|\mathrm{x}\right|\le \sqrt{2}$

Step-by-step solution

Significance of ratio test

The ratio test as follows:

Solution steps:

First,find${\mathbf{\rho }}_{\mathbf{n}}\mathbf{=}\mathbf{\left|}\frac{{\mathbf{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}}{{\mathbf{a}}_{\mathbf{n}}}\mathbf{\right|}\mathbf{.}$.

Second,$\mathbf{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{}{\mathbf{p}}_{\mathbf{n}}\mathbf{.}$.

Third,$\mathbf{p}\mathbf{<}\mathbf{1}$ the series converges, if$\mathbf{p}\mathbf{>}\mathbf{1}$the series diverges.

Finding convergence

Let us consider the power series

$\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{2\mathrm{n}}}{2^{\mathrm{n}}{\mathrm{n}}^2}$

And I want to find the convergence interval of the power series.

Use the ratio test as follows:

Solution steps:

First,find ${\mathrm{\rho }}_{\mathrm{n}}=\left|\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}\right|$.

Second,$\mathrm{\rho }=\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{\rho }}_{\mathrm{n}}$.

Third, $\rho <1$the series converges, if $\rho >1$the series diverges.

localid="1658817382567" $$\begin{gathered} {\mathrm{\rho }}_{\mathrm{n}}=\left|\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}\right| \\ =\left|\frac{{\displaystyle \frac{{\mathrm{x}}^{2\mathrm{n}+2}}{{\left(\mathrm{n}+1\right)}^2{2}^{\mathrm{n}+1}}}}{{\displaystyle \frac{{\mathrm{x}}^{2\mathrm{n}}}{{\mathrm{n}}^2{2}^{\mathrm{n}}}}}\right| \\ =\left|\frac{{\mathrm{x}}^{2\mathrm{n}+2}{\mathrm{n}}^2{2}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2{2}^{\mathrm{n}+1}{\mathrm{x}}^{2\mathrm{n}}}\right| \\ =\left|\frac{{\mathrm{n}}^2{\mathrm{x}}^2}{2{\left(\mathrm{n}+1\right)}^2}\right| \\ \mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{{\mathrm{n}}^2{\mathrm{x}}^2}{2{(\mathrm{n}+1)}^2}\right| \\ =\left|\frac{{\mathrm{x}}^2}{2{\left(1+{\displaystyle \frac{1}{\infty }}\right)}^2}\right| \\ =\left|\frac{{\mathrm{x}}^2}{2}\right| \end{gathered}$$

Finding the points of divergence

The series converges for, $\rho <1$which happens when,$\left|\frac{x^2}{2}\right|<1$and it diverges for $\frac{{\mathrm{x}}^2}{2}>1$.

For the endpoints of the convergence interval, when, $x=\sqrt{2}$the series becomes:

localid="1658819776094" $\sum _{\mathrm{n}=1}^{\infty }\frac{1}{{\mathrm{n}}^2}$

This is a convergent series by using the integral test.

For$x=-\sqrt{2}$ , the series becomes:

$\sum _{\mathrm{n}=1}^{\infty }\frac{1}{{\mathrm{n}}^2}$

This is also a convergent by using the integral test.

Concluding statement 

The convergence is between the intervals$\left|\mathrm{x}\right|\le \sqrt{2}$