Chapter 8: Q. 8 (page 669)

Show that $\sum_{k = 0}^{\infty} \frac{1}{1 + 2^{k}} x^{k}$ , the power series in $x$ from Example 1, diverges when $x = - 2$

Short Answer

Expert verified

Ans: The power series $\sum_{k = 0}^{\infty} \frac{1}{1 + 2^{k}} x^{k}$ diverges when $x = - 2$ .

Step by step solution

Step 1. Given information.

given,

$\sum_{k = 0}^{\infty} \frac{1}{1 + 2^{k}} x^{k}$

Step 2. We evaluate the series when x=-2

So,

$\sum_{k = 0}^{\infty} \frac{1}{1 + 2^{k}} (- 2)^{k} = \sum_{k = 0}^{\infty} \frac{(- 1)^{k} 2^{k}}{1 + 2^{k}}$

Therefore, when $k \to \infty, lim_{x \to \infty} \frac{(- 1)^{k} 2^{k}}{1 + 2^{k}} = 1$

Thus by the divergence test, the power series $\sum_{k = 0}^{\infty} \frac{1}{1 + 2^{k}} x^{k}$ diverges when $x = - 2$

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Show that $\sum_{k = 0}^{\infty} \frac{1}{1 + 2^{k}} x^{k}$ , the power series in $x$ from Example 1, diverges when $x = - 2$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. We evaluate the series when x=-2

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Step by step solution

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Step 2. We evaluate the series when x=-2

One App. One Place for Learning.

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Show that $\sum_{k = 0}^{\infty} \frac{1}{1 + 2^{k}} x^{k}$ , the power series in $x$ from Example 1, diverges when $x = - 2$