Q 58. - 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  $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k}k!}{k^2}{\left(\frac{3x}{x-2}\right)}^k$

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

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

Since the series contain power $\frac{3x}{x-2}$. So the series is not a power series.

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

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

So, by the ratio of absolute convergence, the series will converge when

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

This implies that 

$$\begin{gathered} -1<\frac{3x}{x-2}<1 \\ So, \\ -\left(x-2\right)>3x and 3x>x-2 \\ -x+2>3x and 2x>-2 \\ x<\frac{1}{2} and x>-1 \end{gathered}$$

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