Question

Find the interval of convergence for power series:$\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$

Short answer

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

Step-by-step solution

Step 1. Given information. 

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

Step 2. Find the interval of convergence.

Let us assume $b_k=\frac{1}{k!}{x}^k$androle="math" localid="1649232298108" $b_{k+1}=\frac{1}{\left(k+1\right)!}{x}^{k+1}$

Ratio for the absolute convergence is

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

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$\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$is R.