Question

In Problems 1-10 find the interval and radius of convergence for the given power series.

$\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}}{{\mathbf{9}}^{\mathbf{n}}}{\mathit{x}}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}$

Short answer

The radius of convergence is R = 3 & interval of convergence (-3,3).

Step-by-step solution

Ratio test

The Ratio test for the power series, $\sum _{n=1}^{\infty }{a}_n(x-a{)}^n$is given as,

$\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}{(x-a)}^{n+1}}{a_{n{(\times -a)}^n}}\right|=L$

If$L<1$, then the series converges absolutely.

If$L>1$, then the series diverges.

If$L=1$, then the test is inconclusive.

Apply ratio test

Consider the series$\sum _{k=1}^{\infty }\frac{{(-1)}^n}{9^n}{x}^{2n+1}$

Let$n^{th}$ theterm of the series is$c_n=\frac{{(-1)}^n}{9^n}{x}^{2n+1} , c_{n+1}=\frac{{(-1)}^{n+1}}{9^{n+1}}{x}^{2(n+1)+1}$

The ratio test gives,

$$\begin{gathered} \underset{n\to \infty }{\lim }\left|\frac{{(-1)}^{n+1}{x}^{2(n+1)+1}}{9^{n+1}}\right|\frac{{(-1)}^n{x}^{2n+1}}{9^n}\left[\underset{n\to \infty }{\lim }\right]\frac{{(-1)}^{n+1}{x}^{2(n+1)+1}}{9^{n+1}}·\frac{9^n}{{(-1)}^n{x}^{2n+1}\mid }\mid \\ =\underset{n\to \infty }{\lim }\left|\frac{{(-1)}^n(-1)x^{2n}{x}^3}{{(-1)}^n{x}^{2n}·x}·\frac{9^n}{9^n·9}\right| \\ =\underset{n\to \infty }{\lim }\left|\frac{(-1)x^3}{9x}\right| \\ =\frac{1}{9}\left|x^2\right| \end{gathered}$$

Find interval & radius of convergence

To determine the interval and radius of convergence,

If $\frac{1}{9}\left|x^2\right|<1$

$$\begin{gathered} \left|x^2\right|<9 \\ \left|x\right|<3 \end{gathered}$$

The inequality $\left|x\right|<3$shows that the power series is centered at$a=0$.

And the radius of convergence R = 3.

Now, the interval of convergence is$(a-R,a+R)=(0-3,0+3)$.

Thus, the interval of convergence is$(-3,3)$.

Check convergence at end points

The endpoints are$x=-3$and x=3.

At x = -3, the series becomes

$$\begin{gathered} \sum \frac{{(-1)}^n}{9^n}(-3{)}^{2n+1}=\sum _{n=1}^{\infty }{\left(\frac{-1}{9}\right)}^n{\left({(-3)}^2\right)}^n·(-3) \\ =-3\sum _{n=1}^{\infty }{\left(\frac{-1\left(9\right)}{9}\right)}^k \\ =-3\sum _{n=1}^{\infty }{(-1)}^k\text{diverge}s \end{gathered}$$

At x =3, the series becomes

$$\begin{gathered} \sum _{n=1}^{\infty }\frac{{(-1)}^n}{9^n}(3{)}^{2n+1}=\sum _{n=1}^{\infty }{\left(\frac{-1}{9}\right)}^n{\left({\left(3\right)}^2\right)}^n·\left(3\right) \\ =3\sum _{n=1}^{\infty }{\left(\frac{-1·9}{9}\right)}^k \\ =3\sum _{n=1}^{\infty }{(-1)}^k \end{gathered}$$

Divergent. So, the power series is divergent at both endpoints.

Radius of convergence R = 3 & (-3,3) interval of convergence.