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Chapter 8: Q. 6 (page 692)

Explain why $\lim_{n \to \infty} \frac{{|x|}^{n}}{n!} = 0$ for every value of x.

Short Answer

Expert verified

Using ratio test of convergence,

$\lim_{n \to \infty} \frac{{|x|}^{n + 1}}{n + 1!} \times \frac{n!}{{|x|}^{n}} = \lim_{n \to \infty} |x| \frac{n}{n + 1} = 0$

Step by step solution

01

Step 1. Given information.

The limit is $\lim_{n \to \infty} \frac{{|x|}^{n}}{n!} = 0$ .

02

Step 2. Explanation.

Use the ratio test of convergence.

Let $b_{n} = \frac{{|x|}^{n}}{n!}$ and $b_{k + 1} = \frac{{|x|}^{n + 1}}{n + 1!}$

role="math" localid="1649610457627" $\lim_{n \to \infty} \frac{{|x|}^{n + 1}}{n + 1!} \times \frac{n!}{{|x|}^{n}} = \lim_{n \to \infty} |x| \frac{n}{n + 1} = 0$

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Most popular questions from this chapter

The second-order differential equation

$x^{2} y^{''} + x y^{'} + (x^{2} - p^{2}) = 0$

where p is a non-negative integer, arises in many applications in physics and engineering, including one model for the vibration of a beaten drum. The solution of the differential equation is called the Bessel function of order p, denoted by $J_{p} (x)$ .It may be shown that $J_{p} (x)$ is given by the following power series in x:

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

What is the interval of convergence for $J_{p} (x)$ where p is a non-negative integer

Prove that if the power series $\sum_{k = 0}^{\infty} a_{k} x^{k}$ has a positive and finite radius of convergence $p$ , then the series $\sum_{k = 0}^{\infty} a_{k} x^{2 k}$ has a radius of convergence $\sqrt{p}$ .

Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.

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

In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_{4} (x)$

$\cos x$

Find the interval of convergence for power series: $\sum_{k = 1}^{\infty} {(- 1)}^{K} \frac{1}{k^{k}} {(x - 1)}^{k}$

See all solutions

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