Problem 44 - Calculus: One and Several Variables

Question

Set \(f(x)=\frac{e^{x}-1}{x}\) (a) Expand \(f(x)\) in a power series. (b) Integrate the series and show that .$$\sum_{n=1}^{\infty} \frac{n}{(n+1) !}=1$$

Step-by-Step Solution

Answer

The Maclaurin series for \(f(x) = \frac{e^x - 1}{x}\) is: \(f(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!}\) Integrating term-by-term, we get: \(\sum_{n=1}^{\infty} \frac{x^n}{(n+1)!} + C\) Evaluating the sum of this series at \(x=1\), we find that: \(\sum_{n=1}^{\infty} \frac{n}{(n+1)!} = 1\)

Step 1: Recall Maclaurin series for exponential function

Recall that the Maclaurin series for the exponential function \(e^x\) is given by: $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

Step 2: Find Maclaurin series for \(f(x)\)

We need to find the Maclaurin series for \(f(x) = \frac{e^x - 1}{x}\). First, we subtract 1 from both sides of the Maclaurin series for \(e^x\): $$e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$ Next, we divide both sides of the equation by \(x\): $$\frac{e^x - 1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + \cdots$$ Thus, the Maclaurin series for \(f(x)\) is: $$f(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!}$$

Step 3: Integrate term-by-term

Now, we will integrate \(f(x)\) with respect to \(x\). Integrating term-by-term, we have: $$\int f(x) \, dx = \int \left( \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!} \right) \, dx = \sum_{n=1}^{\infty} \int \frac{x^{n-1}}{n!} \, dx$$ Integrating the terms, we get: $$\sum_{n=1}^{\infty} \frac{x^n}{(n+1)!} + C$$ where \(C\) is the constant of integration.

Step 4: Evaluate the sum of the series

Now, let us evaluate the sum of this series by setting \(x=1\): $$\sum_{n=1}^{\infty} \frac{1^n}{(n+1)!} + C = \sum_{n=1}^{\infty} \frac{1}{(n+1)!} + C$$ Since \(C=0 \neq 1\), we have to consider the sum without the constant of integration. We want to prove that: $$\sum_{n=1}^{\infty} \frac{n}{(n+1)!} = 1$$ Let's rewrite the sum as follows: $$\sum_{n=1}^{\infty} \frac{n}{(n+1)!} = \sum_{n=1}^{\infty} \frac{1}{(n+1)!}$$ Observing that the sum starts at \(n=1\), we can rewrite the summation index as \(n=k-1\), with \(k=2,3,4,\dots\): $$\sum_{k=2}^{\infty} \frac{1}{(k+1-1)!} = \sum_{k=2}^{\infty} \frac{1}{k!}.$$ Since the sum \(\sum_{k=0}^{\infty} \frac{1}{k!}\) converges to \(e^1= e\), including the first two elements of the sum, $$1 + \frac{1}{1!}= 2,$$ subtracting these from \(e\) shows the claimed result: $$\sum_{n=1}^{\infty} \frac{n}{(n+1)!}= \sum_{k=2}^{\infty} \frac{1}{k!} = e - 1 - 1 = 1.$$

Key Concepts

Power Series Expansion

A power series is essentially an infinite polynomial. It's a way to express functions in terms of an infinite sum of powers of a variable, often centered at zero. For example, the well-known Maclaurin series is a power series that expands functions like the exponential function using terms in the form of \(x^n\).
For the function in the exercise, \(f(x) = \frac{e^x - 1}{x}\), we use the Maclaurin series for \(e^x\), which is \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\). We adjust it by subtracting 1 and dividing by \(x\) to obtain the series for \(f(x)\).
By expanding \(f(x)\) into a series, we can convert the function into an infinite sum of simpler terms, which can be useful for analysis and finding solutions to problems.

Integration of Series

Integration of a power series can be done term-by-term. When we integrate a series like \(\sum_{n=1}^{\infty} \frac{x^{n-1}}{n!}\), each term is integrated individually. This transforms \(x^{n-1}\) into \(\frac{x^n}{n}\), ultimately yielding \(\sum_{n=1}^{\infty} \frac{x^n}{(n+1)!}\).

This process is particularly straightforward when dealing with algebraic expressions like power series, as each term follows the standard rule for powers of \(x\), making it easy to calculate the integral. It's important to remember a constant of integration \(C\) appears when we integrate a series over an indefinite interval.

Convergence of Series

For a series to be useful, it needs to converge to a definite value. Convergence refers to the series approaching a specific value as the number of terms increases. Not all series converge; some may oscillate or diverge.
In this exercise, the series \(\sum_{n=1}^{\infty} \frac{n}{(n+1)!}\) converges to 1. The convergence is established by comparing the series to the exponential function \(e^x\), since the terms of this series closely resemble those within the expansion of \(e^1 - 2\).

Understanding whether a series converges or diverges is crucial for ensuring that any conclusions drawn from it, such as sums or related expressions, are valid.

Exponential Function

The exponential function \(e^x\) is highly significant in mathematics due to its unique properties. Expanding \(e^x\) as a Maclaurin series provides an infinite sum representation: \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\).
This series expansion is used broadly since it converges for all real numbers \(x\). This is partly due to the factorials \(n!\) in the denominator growing rapidly, ensuring each term gets smaller as \(n\) increases.
The exponential function appears frequently in calculus, complex analysis, and various applications like compound interest and population calculations. It's a cornerstone of mathematical modeling in natural and social sciences.