/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 Find the sum of the series. $$... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 79

Find the sum of the series. $$ 3+\frac{9}{2 !}+\frac{27}{3 !}+\frac{81}{4 !}+\cdots $$

Short Answer

Expert verified
The sum of the series is approximately 19.0855.

Step by step solution

01

Identify the pattern

Recognize the pattern in the series: each term is given by \( \frac{3^n}{n!} \) where \( n \) starts from 1 and increases by 1 in each subsequent term. Therefore, the series can be written as \( \sum_{n=1}^{ ext{infinity}} \frac{3^n}{n!} \).
02

Recognize the series as exponential

Notice that the series \( \sum_{n=0}^{ ext{infinity}} \frac{a^n}{n!} \) is the Taylor series expansion for the exponential function \( e^a \). In this problem, \( a = 3 \), and the sum starts from \( n=1 \) instead of \( n=0 \).
03

Modify the exponential series

Since the series starts from \( n=1 \), we adjust the exponential series by subtracting the term for \( n=0 \). This gives us \( e^3 - 1 \).
04

Write the final expression for the sum

Combine the findings: the sum of the series is \( S = e^3 - 1 \).
05

Calculate the numerical value

Compute the numerical value of the sum by finding \( e^3 \) approximately using a calculator, which is about 20.0855. So, \( S \approx 20.0855 - 1 = 19.0855 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor series expansion
The Taylor series expansion is a handy tool in mathematics. It allows us to represent complex functions in simpler forms using polynomials. When we expand a function into a Taylor series, we're essentially expressing it as an infinite sum of terms. Each term is calculated using the derivatives of the function at a specific point.
  • The general form of a Taylor series for a function \( f(x) \) about a point \( a \) is: \[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots \]
  • This series can continue indefinitely, hence the name 'infinite series'.
  • Each term in the series becomes smaller and smaller as the powers of \((x-a)\) increase.
  • Taylor series are beneficial because they let us approximate functions that are often too complicated to compute directly.
In our original problem, we utilize the Taylor series expansion to recognize that our series is of the exponential form. Knowing this helps us simplify and solve the series faster using the properties of exponential functions.
Exponential function
The exponential function is a crucial concept in mathematics. It is defined as \( e^x \), where \( e \) is a constant approximately equal to 2.71828. The exponential function appears frequently in calculus due to its unique property: its derivative is the same at any point \( x \), meaning it grows at the same rate it is expressed.
  • The exponential function is significant in many fields like population growth, compound interest, and probability theory.
  • In terms of series, it can be represented by the infinite sum: \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
  • This infinite summation is what we call a Taylor series when centered at zero for the exponential function.
In the context of solving the series from the exercise, recognizing the series as part of the exponential function helped simplify the series into a more manageable form, \( e^3 \), and further adjusting to find the sum excluding the first few terms.
Infinite series
An infinite series is a sum of an infinite sequence of terms. Convergence, or the idea that the series approaches a specific number as more terms are added, is key to understanding infinite series.
  • An infinite series can be both geometric and arithmetic depending on the terms involved.
  • The terms in an infinite series are part of an arithmetic or geometric sequence, but the series itself goes on indefinitely.
However, not all infinite series converge. A series must converge to be meaningful in mathematics. For instance,
  • The series \( \sum_{n=1}^{\infty} \frac{3^n}{n!} \) converges because its terms get smaller, approaching zero as \( n \) increases.
  • In contrast, a divergent series is one where the sum does not approach a finite limit.
Tackling an infinite series like in our exercise by relating it to known, convergent forms such as the exponential series, allows us to handle these infinite sums effectively and compute their values.

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Most popular questions from this chapter

Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number \(a\). (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x)=T_{x}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\). $$ f(x)=x^{2 / 3}, \quad a=1, \quad n=3, \quad 0.8 \leqslant x \leqslant 1.2 $$

Evaluate the indefinite integral as an infinite series. $$ \int x^{2} \sin \left(x^{2}\right) d x $$

Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number \(a\). (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x)=T_{x}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\). $$ f(x)=\sec x, \quad a=0, \quad n=2, \quad-0.2 \leqslant x \leqslant 0.2 $$

Find the Taylor series for \(f(x)\) centered at the given value of \(a\). [ Assume that \(f\) has a power series expansion. Do not show that \(\left.R_{n}(x) \rightarrow 0 .\right]\) Also find the associated radius of convergence. $$ f(x)=\cos x, \quad a=\pi / 2 $$

Find the sum of the series. $$ 1-\ln 2+\frac{(\ln 2)^{2}}{2 !}-\frac{(\ln 2)^{3}}{3 !}+\cdots $$
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