91Ó°ÊÓ

The ratio test as follows:

Solution steps:

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

Second,$\mathbf{ÒÏ}\mathbf{=}\underset{\mathbf{n}\mathbf{→}\mathbf{∞}}{\mathbf{lim}}\mathbf{}{\mathbf{p}}_{\mathbf{n}}\mathbf{.}$.

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

Let us consider the power series

$\underset{\mathrm{n}=1}{\overset{∞}{∑}}\frac{{\mathrm{x}}^{2\mathrm{n}}}{2^{\mathrm{n}}{\mathrm{n}}^2}$

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

Use the ratio test as follows:

Solution steps:

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

Second,$\mathrm{ÒÏ}=\underset{\mathrm{n}→∞}{\lim }{\mathrm{ÒÏ}}_{\mathrm{n}}$.

Third, $\mathrm{ÒÏ}<1$the series converges, if $\mathrm{ÒÏ}>1$the series diverges.

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

The series converges for, $\mathrm{ÒÏ}<1$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" $\underset{\mathrm{n}=1}{\overset{∞}{∑}}\frac{1}{{\mathrm{n}}^2}$

This is a convergent series by using the integral test.

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

$\underset{\mathrm{n}=1}{\overset{∞}{∑}}\frac{1}{{\mathrm{n}}^2}$

This is also a convergent by using the integral test.

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