Chapter 8: Q. 69 (page 671)

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

Short Answer

Expert verified

Ans: It is proved that the radius of convergence of the power series $\sum_{k = 0}^{\infty} b^{k} x^{k} is \frac{1}{| b |}$

Step by step solution

Step 1. Given information.

given,

$\sum_{k = 0}^{\infty} b^{k} x^{k} is \frac{1}{| b |}$

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

Also, let us consider $b_{k} = a_{k} x^{k}, so b_{k + 1} = a_{k + 1} x^{k + 1}$

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

$\begin{aligned} lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = lim_{k \to \infty} |\frac{b^{k + 1}}{b^{k}} x| \\ = lim_{k \to \infty} | b x | \end{aligned}$

So according to the ratio test for absolute convergence, the series will converge only when $| b x | < 1$

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

Step 3. Thus,

The radius of convergence of the power series $\sum_{k = 0}^{\infty} b^{k} x^{k} is \frac{1}{| b |}$

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

Short Answer

Step by step solution

Step 1. Given information.

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

Step 3. Thus,

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Let bbe a nonzero constant. Prove that the radius of convergence of the power series ∑k=0∞ bkxkis1|b|.

Short Answer

Step by step solution

Step 1. Given information.

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

Step 3. Thus,

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

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Study anywhere. Anytime. Across all devices.

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