Q 33. - Calculus

Question

Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.

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

Step-by-Step Solution

Answer

The interval of convergence for power series is $\left(-\frac{5}{2},\frac{7}{2}\right)$.

Step 1. Given information.

The given power series is:

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

Step 2. Find the interval of convergence.

Let us assume $b_k={\left(-1\right)}^k\frac{k^2}{3^k}{\left(x-\frac{1}{2}\right)}^k$ therefore

$b_{k+1}={\left(-1\right)}^{k+1}\frac{{\left(k+1\right)}^2}{3^{k+1}}{\left(x-\frac{1}{2}\right)}^{k+1}$

The ratio for the absolute convergence is 

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\left(-1\right)}^{k+1}{\displaystyle \frac{{\left(k+1\right)}^2}{3^{k+1}}}{\left(x-{\displaystyle \frac{1}{2}}\right)}^{k+1}}{{\left(-1\right)}^k{\displaystyle \frac{k^2}{3^k}}{\left(x-{\displaystyle \frac{1}{2}}\right)}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|\left(-1\right){\left(\frac{k+1}{k}\right)}^2\frac{3^k}{3^{k+1}}\left(x-\frac{1}{2}\right)\right| \\ =\underset{k\to \infty }{\lim }\left|x-\frac{1}{2}\right|{\left(\frac{k+1}{k}\right)}^2\frac{1}{3} \end{gathered}$$

Here the limit $\frac{1}{3}\left|x-\frac{1}{2}\right|$ is So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\frac{1}{3}\left|x-\frac{1}{2}\right|<1$ that is $-\frac{5}{2}<x<\frac{7}{2}$.

Step 3. Find the interval of convergence.

Now, since the intervals are finite so we analyze the behavior of the series at the endpoints.

So, when $x=-\frac{5}{2}$

$$\begin{gathered} \sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{k^2}{3^k}{\left(x-\frac{1}{2}\right)}^k=\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{k^2}{3^k}{\left(-\frac{5}{2}-\frac{1}{2}\right)}^k \\ =\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{k^2}{3^k}{\left(-3\right)}^k \\ =\sum _{k=0}^{\infty }{k}^2{\left(1\right)}^k \end{gathered}$$

The result is just a constant multiple of the harmonic series, which diverges.

So, when $x=\frac{7}{2}$

$$\begin{gathered} \sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{k^2}{3^k}{\left(x-\frac{1}{2}\right)}^k=\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{k^2}{3^k}{\left(\frac{7}{2}-\frac{1}{2}\right)}^k \\ =\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{k^2}{3^k}{\left(3\right)}^k \\ =\sum _{k=0}^{\infty }{k}^2{\left(-1\right)}^k \end{gathered}$$

The result is the alternating multiple of the arithmetic series, which diverges conditionally. 

Therefore, the interval of convergence of the power series $\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{k^2}{3^k}{\left(x-\frac{1}{2}\right)}^k$ is $\left(-\frac{5}{2},\frac{7}{2}\right)$.