Question

Find the interval of convergence of the power series

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

Short answer

The interval of convergence of the power series$\sum _{k=1}^{\infty }(x-3{)}^k$is$(2,4)$

Step-by-step solution

Given information  

The power series is$\sum _{k=1}^{\infty }(x-3{)}^k$

The ratio test for absolute convergence will be used to determine the convergence interval.

Let, the first assume, therefore

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

The limit is$|x-3|$

The ratio test for absolute convergence will be used to determine the convergence interval when $|x-3|<1$that is $-1<x-3<1$

As a result, we may write $-1<x-3$and $x-3<1$

Implies that

$x>2andx<4$

or$x\in (2,4)$

Now, because the intervals are limited, we examine the series' behavior at the ends.

When $x=2$

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

$=\sum _{k=1}^{\infty }(-1{)}^k$

The series contains the conditionally convergent alternating multiple.

When $x=4$

$$\begin{gathered} {\left.\sum _{k=1}^{\infty }(x-3{)}^k\right|}_{x=4}=\sum _{k=1}^{\infty }(4-3{)}^k \\ =\sum _{k=1}^{\infty }(1{)}^k \\ =\sum _{k=1}^{\infty }1 \end{gathered}$$

The series is a diverging constant multiple.