Q 24. - Calculus

Question

Find the interval of convergence for power series: $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{x}^{2k+1}$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given power series is $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{x}^{2k+1}$.

Step 2. Find the interval of convergence.

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

Ratio for the absolute convergence is 

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

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

So, $\underset{k\to \infty }{\lim }\left|x^2\right|\frac{-1}{\left(2k+3\right)\left(2k+2\right)}=0$  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 $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{x}^{2k+1}$ is $R$.