Q 57. - Calculus

Question

Explain why the series is not a power series in $x-x_0$.Then use the ratio test for absolute convergence to find the values of $x$ for which the given series converge   $\sum _{k=1}^{\infty }\frac{1}{k^3}{\left(\frac{x+2}{x-3}\right)}^k$

Step-by-Step Solution

Answer

The value of $x$for which the series $\sum _{k=1}^{\infty }\frac{1}{k^3}{\left(\frac{x+2}{x-3}\right)}^k$converges when $\left(1,\infty \right)$.

Step 1. Given information.

The given power series is $\sum _{k=1}^{\infty }\frac{1}{k^3}{\left(\frac{x+2}{x-3}\right)}^k$.

Step 2. Find the values of x for which the given series converge.

Since, the series consists of power of $\frac{x+2}{x-3}$So the series in not power series

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

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

By ratio test of absolute convergence, the series will converge when 

$\left|\frac{x+2}{x-3}\right|<1$

This implies that 

$-1<\frac{x+2}{x-3}<1$

So, $-\left(x-3\right)>x+2$ and $x+2>x-3$

Solve the first inequality

$$\begin{gathered} -x+3>x+2 \\ 2x<1 \\ x<\frac{1}{2} \end{gathered}$$

and the second inequality is invalid.

Therefore, the value of  $x$ for which the series $\sum _{k=1}^{\infty }\frac{1}{k^3}{\left(\frac{x+2}{x-3}\right)}^k$converges when $\left(1,\infty \right)$