Q 54. - 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 $\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{1}{k!}{x}^{-k}$

Step-by-Step Solution

Answer

The value of $x$for which the series  $\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{1}{k!}{x}^{-k}$ converges when $x\in R$.

Step 1. Given information.

The given power series is $\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{1}{k!}{x}^{-k}$.

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

Since, the power of the $x$ is negative. So the series is not power series.

Now, the value of $b_k$ is

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

Thus,

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

So, for $k\to \infty$ the value of limit will always be zero.

Therefore the value of $x$ for which the series  $\sum _{k=0}^{\infty }{\left(-1\right)}^k\frac{1}{k!}{x}^{-k}$ converges when $x\in R$.