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 $\sum_{n - 1}^{\infty} \frac{2^{n}}{n!}$ .

Short Answer

Expert verified

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

Step by step solution

Given Information

The given series, i.e. $\sum_{n - 1}^{\infty} \frac{2^{n}}{n!}$ ,

Definition of Sum of a Series

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

Find the sum of the series

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

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + . . . . e^{x} = \sum_{n - 0}^{\infty} \frac{x^{n}}{n!} = 1 + \sum_{n - 0}^{\infty} \frac{x^{n}}{n!}$

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

$e^{2} = 1 + \sum_{n - 1}^{\infty} \frac{2^{n}}{n!} e^{2} - 1 = \sum_{n - 1}^{\infty} \frac{2^{n}}{n!} \sum_{n - 1}^{\infty} \frac{2^{n}}{n!} = e^{2} - 1$

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

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

Short Answer

Step by step solution

Given Information

Definition of Sum of a Series

Find the sum of the series

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Find the sum of each of the following series by recognizing it as the Maclaurin series for a function evaluated at a point∑n-1∞2nn!.

Short Answer

Step by step solution

Given Information

Definition of Sum of a Series

Find the sum of the series

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Kinematics Physics

Turning Points in Physics

Measurements

Work Energy and Power

Solid State Physics

Energy Physics

Study anywhere. Anytime. Across all devices.

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