Q. 3 - 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.

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

Step-by-Step Solution

Answer

The interval of convergence is $2<x<4$ and the radius of convergence is $R=1$.

Step 1: Given information

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

Step 2: Find the interval of convergence and radius of convergence.

By using the ratio test,

$$\begin{gathered} a_k=\ln k{(x-3)}^k{a}_{k+1}=\ln (k+1){(x-3)}^{k+1}p=\underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right| \\ p=\underset{k\to \infty }{\lim }\left|\frac{xk{(\ln k)}^2}{(k+1){\left[\ln \right(k+1\left)\right]}^2}\right| \\ p=\underset{k\to \infty }{ \lim }\left|\frac{(x-3)\ln (k+1)}{\ln k}\right| \\ p=\underset{k\to \infty }{ \left|x-3\right| \lim }\left|\frac{\ln (k+1)}{\ln k}\right| \\ p=\left|x-3\right|\times 1 \\ p=\left|x-3\right| \end{gathered}$$

The series converges if, $p<1$

Thus, 

$$\begin{gathered} \left|x-3\right|<1 \\ -1<x-3<1 \\ -1+3<x<1+3 \\ 2<x<4 \end{gathered}$$

The interval of convergent is $2<x<4$ and the radius of converges is $R=1$.