Q 22. - Calculus

Question

Find the interval of convergence for power series: $\sum _{k=0}^{\infty }\frac{{\left(-2\right)}^k}{k!}{x}^k$

Step-by-Step Solution

Answer

The interval of convergence for power series is $R$ .

Step 1. Given information.

The given power series is $\sum _{k=0}^{\infty }\frac{{\left(-2\right)}^k}{k!}{x}^k$.

Step 2. Find the interval of convergence.

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

Ratio for the absolute convergence is 

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

Now, we evaluate the limit at $k\to \infty$

So, $\underset{k\to \infty }{\lim }\left|x\right|\frac{1}{k+1}=0$ that is the value of limit will be zero no matter what value the variable $x$ takes.

By ratio test, the series converges absolutely for every value of $x$.

Therefore, the interval of convergence of the power series $\sum _{k=0}^{\infty }\frac{{\left(-2\right)}^k}{k!}{x}^k$ is $R$.