Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{2}\mathbf{x}{\mathbf{)}}^{\mathbf{n}}}{{\mathbf{3}}^{\mathbf{n}}}$

The ratio test as follows:

Solution steps:

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

Second,$\mathit{\rho }\mathbf{=}\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathit{\rho }}_{\mathbf{n}}\mathbf{.}$ .

Third,$\mathbf{}\mathit{\rho }\mathbf{<}\mathbf{1}$ t he series converges, if$\mathbf{}\mathit{\rho }\mathbf{>}\mathbf{1}$the series diverges.

Let us consider the power series:

$\sum _{\mathrm{n}=0}^{\infty }\frac{(2\mathrm{x}{)}^{\mathrm{n}}}{3^{\mathrm{n}}}$

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

Use the ratio test as follows:

Solution steps:

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

Second,$\rho =li{m}_{\mathrm{n}\to \infty }{\rho }_{\mathrm{n}}$.

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

localid="1658741466259" $$\begin{gathered} {\rho }_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}} \\ =\left|\frac{{\displaystyle \frac{{\left(2x\right)}^{n-1}}{3^{n-1}}}}{{\displaystyle \frac{{\left(2x\right)}^n}{3^n}}}\right| \\ {\rho }_{\mathrm{n}}=\left|\frac{3^n{\left(2x\right)}^{n-1}}{{\left(2x\right)}^n{3}^{n-1}}\right| \\ =\left|\frac{2x}{3}\right| \\ \rho =\underset{n-x}{lim}\left|\frac{2x}{3}\right| \\ =\left|\frac{2x}{3}\right| \end{gathered}$$

The series converges for, which happens when $\left|\frac{2x}{3}\right|<1$,and it diverges$\left|\frac{2x}{3}\right|>1$ for

Therefore, the series converges when $-1<\frac{2x}{3}<1$, in another words the series converges when role="math" localid="1658741876573" $\frac{-3}{2}<x<\frac{3}{2}$.

For the endpoints of the convergence interval, whenrole="math" localid="1658741800063" $x=\frac{3}{2}$ , the series becomes:

$\sum _{\mathrm{n}=0}^{\infty }{\left(1\right)}^n$

which is a divergent series.

For t $x=\frac{-3}{2}$, the series becomes:

$\sum _{\mathrm{n}=0}^{\infty }{\left(-1\right)}^n$

which is also a divergent series.

The convergence is between the intervals$\frac{-3}{2}<x<\frac{3}{2}$ .