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Chapter 1: Q16P (page 41)

Find the sum of each of the following series by recognizing it as the Maclaurin series for a function evaluated at a point ${鈭�}_{n - 1}^{鈭�} \frac{2^{n}}{n!}$ .

Short Answer

Expert verified

The sum of the series, i.e. ${鈭�}_{n - 1}^{鈭�} \frac{2^{n}}{n!} = e^{2} - 1$ ,

Step by step solution

01

Given Information

The given series, i.e. ${鈭�}_{n - 1}^{鈭�} \frac{2^{n}}{n!}$ ,

02

Definition of Sum of a Series

The total of all the terms in a sequence can be written as the sum of a series.

03

Find the sum of the series

Use the Maclaurin series of $e^{x}$ as:

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + . . . . e^{x} = {鈭�}_{n - 0}^{鈭�} \frac{x^{n}}{n!} = 1 + {鈭�}_{n - 0}^{鈭�} \frac{x^{n}}{n!}$

Further, put $x = 2$ and simplify, and we get,

$e^{2} = 1 + {鈭�}_{n - 1}^{鈭�} \frac{2^{n}}{n!} e^{2} - 1 = {鈭�}_{n - 1}^{鈭�} \frac{2^{n}}{n!} {鈭�}_{n - 1}^{鈭�} \frac{2^{n}}{n!} = e^{2} - 1$

Hence, the sum of the series, i.e. ${鈭�}_{n - 1}^{鈭�} \frac{2^{n}}{n!} = e^{2} - 1$ ,

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Most popular questions from this chapter

Show thatis the distance between the points and in the complex plane. Use this result to identify the graphs in Problems $53, 54, 59, a n d 60$ without computation.

Find the values of several derivatives of $e^{- 1 / t^{2}}$ at t = 0. Hint:Calculate a few derivatives (as functions of t); then make the substitution $x = 1 / t^{2}$ , and use the result of Problem 24(f) or 25.

Write the Maclaurin series for $\frac{1}{\sqrt{1 + x}}$ in $鈭�$ form using the binomial coefficient notation. Then find a formula for the binomial coefficients in terms ofn as we did in Example $2$ above

$\frac{s i n (\sqrt{x})}{\sqrt{x}}, x > 0$

Show that the binomial coefficients $(\begin{array}{l} - 1 \\ n \end{array}) = (- 1)^{n}$

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