Q 45. - Calculus

Question

Find the interval of convergence for power series: $\sum _{k=1}^{\infty }\frac{1}{2.4.6.......\left(2k\right)}{x}^k$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given power series is $\sum _{k=1}^{\infty }\frac{1}{2.4.6.......\left(2k\right)}{x}^k$.

Step 2. Find the interval of convergence.

Let us assume $b_k=\frac{1}{2.4.6......\left(2k\right)}{x}^k$ therefore $b_{k+1}=\frac{1}{2.4.6.......2\left(k+1\right)}{x}^{k+1}$

Ratio for the absolute convergence is 

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{1}{2.4.6......2\left(k+1\right)}}{x}^{k+1}}{{\displaystyle \frac{1}{2.4.6.....2\left(k\right)}}{x}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{1}{2\left(k+1\right)}x\right| \end{gathered}$$

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

This implies that 

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

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

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