Question

In Problems 1-10 find the interval and radius of convergence for the given power series.$\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathbf{3}}^{\mathbf{-}\mathbf{k}}{\mathbf{(}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{)}}^{\mathbf{k}}$

Short answer

The radius of convergence is$R=\frac{5}{4}$& converges interval is.$(\frac{1}{2},2)$

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,

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

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

Rewrite the given power series as follows:

$\begin{array}{c}\sum _{k=1}^{\infty }{3}^{-k}{\left(4x-5\right)}^k=\sum _{k=1}^{\infty }{\left(3\right)}^{-k}{\left(4\right)}^k{(x-\frac{5}{4})}^k\\ =\sum _{k=1}^{\infty }{\left(\frac{3}{4}\right)}^{-k}{\left(x-\frac{5}{4}\right)}^k\end{array}$

Compare this power series with.$\sum _{k=0}^{\infty }{c}_k{(x-a)}^k$

$c_k={\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^{-k},a=\frac{5}{4}=R\text{and}{c}_k\ne 0.$

$\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}{(x-a)}^{n+1}}{c_n{(x-a)}^n}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\left(\frac{3}{4}\right)}^{-(k+1)}{(x-\frac{5}{4})}^{k+1}\mid }{{\left(\frac{3}{4}\right)}^{-k}{(x-\frac{5}{4})}^k}\right|\\ =\underset{k\to \infty }{\lim }\left|\left(\frac{4}{3}\right)\left(x-\frac{5}{4}\right)\right|\\ =\frac{4}{3}\left|x-\frac{5}{4}\right|\\ =\left|\frac{4x-5}{3}\right|\end{array}$

Interval of convergence

By Ratio test, the power series is convergent when$L<1$

$\left|\frac{{\displaystyle 4x-5}}{{\displaystyle 3}}\right|<1$

Since$\left|f\right(x\left)\right|<a$implies,

$f\left(x\right)<a$or$f\left(x\right)>-a$

Here,$\left|\frac{{\displaystyle 4x-5}}{{\displaystyle 3}}\right|<1$, implies,$\frac{4x-5}{3}<1or\frac{4x-5}{3}>-1$.

$\begin{array}{l}\frac{4x-5}{3}<1\frac{4x-5}{3}>-1\\ 4x-5<34x-5>-3\\ 4x<84x>2\\ x<\frac{2}{1}x>\frac{1}{2}\end{array}$

Combining both the ranges of, the interval of convergence is.$\left(\frac{{\displaystyle 1}}{{\displaystyle 2}},2\right)$

Check convergence at end points

At$x=\frac{1}{2}$;

The power series becomes,

$\begin{array}{c}\sum _{k=1}^{\infty }{3}^{-k}{\left(4\left(\frac{1}{2}\right)-5\right)}^k=\sum _{k=1}^{\infty }{3}^{-k}{\left(-3\right)}^k\\ =\sum _{k=1}^{\infty }{(-1)}^k\end{array}$

At $x=2$;

The power series becomes,

$\begin{array}{c}\sum _{k=1}^{\infty }{3}^{-k}{\left(4\left(2\right)-5\right)}^k=\sum _{k=1}^{\infty }{3}^{-k}{\left(3\right)}^k\\ =\sum _{k=1}^{\infty }1\\ =\infty \end{array}$

At the end points, the power series diverges.

Hence, the radius is $R=\frac{5}{4}$and the interval of convergence is$\left(\frac{{\displaystyle 1}}{{\displaystyle 2}},2\right)$.