Q 56. - 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   $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k}{\left(\frac{x}{x-1}\right)}^k$.

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

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

Since the series contains the power of $\frac{x}{x-1}$.So the series is not a power series.

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

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

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

$\left|\frac{x}{\left(x+1\right)}\right|<1$

This implies that $-1<\frac{x}{x-1}<1$

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

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