Q 46. - Calculus

Question

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

Step-by-Step Solution

Answer

The interval of convergence is \( (-\infty, \infty) \), i.e., \( R = \infty \).

Step 1: Apply the ratio test

\( \left|\frac{a_{k+1}}{a_k}\right| = \left|\frac{x^{k+1}}{1\cdot3\cdot5\cdots(2k+3)} \cdot \frac{1\cdot3\cdot5\cdots(2k+1)}{x^k}\right| = \frac{|x|}{2k+3} \to 0 \) as \( k \to \infty \).

Step 2: Conclusion

Since the limit is 0 for all \( x \), the radius of convergence is \( R = \infty \). The interval of convergence is \( (-\infty, \infty) \).