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Chapter 8: Q 43. (page 670)

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{\sqrt{k}}{k!} {(4 x + 7)}^{k}$

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given information.

The given power series is ${鈭�}_{k = 0}^{鈭�} \frac{\sqrt{k}}{k!} {(4 x + 7)}^{k}$ .

02

Step 2. Find the interval of convergence.

Let us assume $b_{k} = \frac{\sqrt{k}}{k!} {(4 x + 7)}^{k}$ and $b_{k + 1} = \frac{\sqrt{k + 1}}{(k + 1)!} {(4 x + 7)}^{k + 1}$

Ratio for the absolute convergence is

$\lim_{k 鈫� 鈭�} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k 鈫� 鈭�} |\frac{\frac{\sqrt{k + 1}}{(k + 1)!} {(4 x + 7)}^{k + 1}}{\frac{\sqrt{k}}{k!} {(4 x + 7)}^{k}}| = \lim_{k 鈫� 鈭�} |4 x + 7| \sqrt{\frac{k + 1}{k} \frac{1}{k + 1}} = \lim_{k 鈫� 鈭�} |4 x + 7| \sqrt{\frac{1}{k (k + 1)}}$

Now, we evaluate the limit at $k 鈫� 鈭�$ .

So, $\lim_{k 鈫� 鈭�} |4 x + 7| \sqrt{\frac{1}{k (k + 1)}} = 0$ , that is the value of limit will be zero no matter what value the variable $x$ takes.

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

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

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Most popular questions from this chapter

Find the interval of convergence for each power series in Exercises 21鈥�48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.

${鈭�}_{k = 0}^{鈭�} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$

If m is a positive integer, how can we find the Maclaurin series for the function $f (c x^{m})$ if we already know the Maclaurin series for the function f(x)? How do you find the interval of convergence for the new series?

Complete Example 4 by showing that the power series ${鈭�}_{k = 0}^{鈭�} 鈥� (鈭� 1)^{k} \frac{k}{3 k + 1} (2 x 鈭� 3)^{k}$ diverges when $x = 2$ .

Let $f$ be a function with an nth-order derivative at a point $x_{o}$ and let $P_{n} (x) = {鈭�}_{k = 0}^{n} \frac{f^{(k)} (x_{0})}{k!} {(x - x_{0})}^{k}$ . Prove that $f^{(k)} (x_{0}) = P_{n}^{(k)} (x_{0})$ for every non-negative integer $k 鈮� n$ .

Show that the power series ${鈭�}_{k = 1}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{k} x^{k}$ converges conditionally when $x = 1$ and diverges when $x = - 1$ . What does this behavior tell you about the interval of convergence for the series?

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