Q. 7 - Calculus

Question

Interval of convergence and radius of convergence: Find the interval of convergence and radius of convergence for each of the given power series. If the interval of convergence is finite, test the series for convergence at each of the endpoints of the interval.

\(\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k2^{k}}\left ( x-2 \right )^{k}\)

Step-by-Step Solution

Answer

The interval of convergence and radius of convergence for each of the given power series is \((0,4]\) and 2.

Step 1. Given Information

The given power series is \(\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k2^{k}}\left ( x-2 \right )^{k}\).

We have to find the interval of convergence and radius of convergence for each of the given power series and If the interval of convergence is finite then we have to test the series for convergence at each of the endpoints of the interval. 

Step 2. Find the interval of convergence

We will use the ratio test for absolute convergence to determine the interval of convergence for the series.

Now, let \(a_{k}= \frac{\left ( -1 \right )^{k}}{k2^{k}}\left ( x-2 \right )^{k} so, a_{k+1}=\frac{\left ( -1 \right )^{k+1}}{\left ( k+1 \right )2^{k+1}}\left ( x-2 \right )^{k+1}\).

So,

\(\begin{aligned}\lim _{k \rightarrow \infty}\left|\frac{a_{k+1}}{a_k}\right| &=\lim _{k \rightarrow \infty}\left|\frac{\frac{\left ( -1 \right )^{k+1}}{\left ( k+1 \right )2^{k+1}}\left ( x-2 \right )^{k+1}}{\frac{\left ( -1 \right )^{k}}{k2^{k}}\left ( x-2 \right )^{k}}\right| \\&=\lim _{k \rightarrow \infty}\frac{1}{2}\frac{k}{\left ( k+1 \right )}\left|x-2 \right|\end{aligned}\)

Now, the limit is \(\frac{1}{2}\left|x-2 \right|\) . So. by the ratio test of absolute convergence, the series will converge.

So, 

\(\frac{1}{2}\left|x-2 \right|< 1\)

\(-2< x-2< 2\)

\(-2< x-2\) and \(x-2< 2\)

\(x> 0\) and \(x< 4\)

\(x\varepsilon \left ( 0,4 \right )\)

Step 3. Find the interval of convergence

Since the intervals are finite we have to test the series for convergence at each of the endpoints of the interval. 

When \(x=0\)

\(\begin{aligned}\left.\sum_{k=1}^{\infty} \frac{(-1)^k}{k 2^k}(x-2)^k\right|_{x=0} &=\sum_{k=1}^{\infty} \frac{(-1)^k}{k 2^k}(0-2)^k \\&=\sum_{k=1}^{\infty} \frac{(-1)^k}{k 2^k}(-1)^k 2^k \\&=\sum_{k=1}^{\infty} \frac{1}{k}(-1)^{2 k} \\&=\sum_{k=1}^{\infty} \frac{1}{k}\end{aligned}\)

The series has alternating multiple which diverges.

When \(x=4\)

The series has alternating multiple which converge.

The interval of convergence of the given power series is \((0,4]\).

The radius of convergence for each of the given power series is 2.