Q. 69 - Calculus

Question

 Let $b$ be a nonzero constant. Prove that the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} b^k{x}^k\text{ is }\frac{1}{\left|b\right|}$. 

Step-by-Step Solution

Answer

Ans:  It is proved that the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} b^k{x}^k\text{ is }\frac{1}{\left|b\right|}$

Step 1. Given information.

given,

     $\sum _{k=0}^{\mathrm{\infty }} b^k{x}^k\text{ is }\frac{1}{\left|b\right|}$

Step 2. Consider the power series ∑ k = 0 ∞   b k x k   , where b be a non-zero constant.

    Also, let us consider $b_k=a_k{x}^k,\text{ so }{b}_{k+1}=a_{k+1}{x}^{k+1}$

Apply the ratio test for absolute convergence in the power series $\sum _{k=0}^{\mathrm{\infty }} b^k{x}^k$, that is

     $\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b^{k+1}}{b^k}x\right|\\ =\underset{k\to \mathrm{\infty }}{lim} \left|bx\right| \end{array}$

So according to the ratio test for absolute convergence, the series will converge only when $\left|bx\right|<1$

Implies that $\left|x\right|<\frac{1}{\left|b\right|}$

Step 3. Thus,

The radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} b^k{x}^k\text{ is }\frac{1}{\left|b\right|}$