Question

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

$\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{k}}}{{\mathbf{k}}^{\mathbf{2}}\mathbf{+}\mathbf{k}}$

Short answer

The radius of convergence is$R=\frac{1}{3}$& converges interval is$\left[0,\frac{{\displaystyle 2}}{{\displaystyle 3}}\right]$.

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

Consider the series.$\sum _{k=1}^{\infty }\frac{{(3x-1)}^k}{k^2+k}$

The Ratio test gives:

$\begin{array}{c}\underset{k\to \infty }{\lim }\left|\frac{\frac{{(3x-1)}^{k+1}}{{(k+1)}^2+k+1}}{\frac{{(3x-1)}^k}{k^2+k}}\right|=\underset{k\to \infty }{\lim }\left|\frac{{(3x-1)}^{k+1}}{{(k+1)}^2+k+1}\cdot \frac{k^2+k}{{(3x-1)}^k}\right|\\ =\underset{k\to \infty }{\lim }\left|\frac{k^2+k}{k^2+3k+2}\cdot \frac{{(3x-1)}^k(3x-1)}{{(3x-1)}^k}\right|\\ =\underset{k\to \infty }{\lim }\left|\frac{k^2+k}{k^2+3k+2}\cdot \frac{3x-1}{1}\right|\\ =|3x-1|\\ =3|x-\frac{1}{3}|\end{array}$

check for convergence

For series converges absolutely

$\begin{array}{c}\text{If}3|x-\frac{1}{3}|<1\\ \Rightarrow |x-\frac{1}{3}|<\frac{1}{3}\\ \text{Hereradiusofconvergence}R=\frac{1}{3}\text{and}a=\frac{1}{3}\\ \Rightarrow 0<x<\frac{2}{3}(\text{since}a-R<a<a+R)\end{array}$

Find the convergence at the left end point of the open interval.$\left(0,\frac{{\displaystyle 2}}{{\displaystyle 3}}\right)$

Let$x=0$

$\sum _{k=1}^{\infty }\frac{{\left(3\right(0)-1)}^k}{k^2+k}=\sum _{k=1}^{\infty }\frac{{(-1)}^k}{k^2+k}$

Now we have to determine the convergence of the series using alternating series test.

Apply alternating series test

Now the alternating series test$\sum _{k=1}^{\infty }\frac{{\left(3\right(0)-1)}^k}{k^2+k}=\sum _{k=1}^{\infty }\frac{{(-1)}^k}{k^2+k}$

Let $u_k=\frac{1}{k^2+k}$& clearly is decreasing and

$\begin{array}{c}\underset{k\to \infty }{\lim }{u}_k=\underset{k\to \infty }{\lim }\frac{1}{k^2+k}\\ =\frac{1}{\infty }\\ =0\end{array}$

Hence, the series$\sum _{k=1}^{\infty }\frac{{(-1)}^k}{k^2+k}$ is converges.

Therefore, the series $\sum _{k=1}^{\infty }\frac{{(3x-1)}^k}{k^2+k}$is converges at $x=0$.

Find the convergence at the right end point of the open interval $\left(0,\frac{{\displaystyle 2}}{{\displaystyle 3}}\right)$.

Find interval & radius of convergence

Check for$x=\frac{2}{3}$

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

So, it is convergent by comparison test.

Let$u_k=\frac{1}{k^2+k}$and$v_k=\frac{1}{k^2}$

Now consider $\underset{k\to \infty }{\lim }\frac{u_k}{v_k}$

$\begin{array}{c}\underset{k\to \infty }{\lim }\frac{u_k}{v_k}=\underset{k\to \infty }{\lim }\frac{\frac{1}{k^2+k}}{\frac{1}{k^2}}\\ =\underset{k\to \infty }{\lim }\frac{k^2}{k^2+k}\\ =1\text{(Finiteandpositive)}\end{array}$

Therefore, both the series $\sum u_n$and,$\sum v_n$ either convergent or divergent.

Hence, the series$\sum _{k=1}^{\infty }\frac{{(3x-1)}^k}{k^2+k}$ is convergent at.$x=\frac{2}{3}$

Hence, the radius and interval of convergenceis $R=\frac{1}{3}$and interval is $\left[0,\frac{{\displaystyle 2}}{{\displaystyle 3}}\right]$.