Question

Find the interval of convergence for power series:$\sum _{k=0}^{\infty }\frac{1}{1.3.5.....\left(2k+1\right)}{x}^k$

Short answer

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

Step-by-step solution

Step 2. Find the interval of convergence.  

Let us assume $b_k=\frac{1}{1.3.5......\left(2k+1\right)}{x}^k$therefore $b_{k+1}=\frac{1}{1.3.5.....\left(2\left(k+1\right)+1\right)}{x}^{k+1}$

Ratio for the absolute convergence is

$\underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|$=

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{\frac{1}{1.3.5.....\left(2k+2+1\right)}{x}^{k+1}}{\frac{1}{1.3.5.....\left(2k+1\right)}{x}^k}\right|=\underset{k\to \infty }{\lim }\left|\frac{\frac{1}{1.3.5.....\left(2k+3\right)}{x}^{k+1}}{\frac{1}{1.3.5.....\left(2k+1\right)}{x}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{1}{2k+3}x\right| \end{gathered}$$

Since for $k\to \infty$the limit is zero irrespective of the value of variable.

This implies that

$\underset{k\to \infty }{\lim }\left|\frac{1}{2k+3}x\right|=0$

Hence by the ratio test the series converges absolutely for every value of $x$.

Therefore, the interval of convergence of the power series is$R$.