Evaluating an infinite series Let f(x) = (eˣ − 1)/x, for x ≠ 0, and f(0)=1. Use the Taylor series for f centered at 0 to evaluate f(1) and to find the value of ∑ₖ₌₀∞ 1/(k+1)!

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Recognize that the function is given by \(f(x) = \frac{e^{x} - 1}{x}\) for \(x \neq 0\), and \(f(0) = 1\). Our goal is to find the Taylor series of \(f\) centered at 0, then use it to evaluate \(f(1)\) and the infinite sum \(\sum_{k=0}^{\infty} \frac{1}{(k+1)!}\).

Recall the Taylor series expansion of \(e^{x}\) centered at 0: \(e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}\). Substitute this into the expression for \(f(x)\) to write \(f(x)\) as a series:

\[f(x) = \frac{e^{x} - 1}{x} = \frac{\sum_{n=0}^{\infty} \frac{x^{n}}{n!} - 1}{x} = \frac{\sum_{n=1}^{\infty} \frac{x^{n}}{n!}}{x}.\]

Simplify the fraction by dividing each term in the numerator by \(x\): \(f(x) = \sum_{n=1}^{\infty} \frac{x^{n}}{n!} \cdot \frac{1}{x} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!}\). Reindex the sum by letting \(k = n - 1\), so that:

\[f(x) = \sum_{k=0}^{\infty} \frac{x^{k}}{(k+1)!}.\]

To find \(f(1)\), substitute \(x=1\) into the series: \(f(1) = \sum_{k=0}^{\infty} \frac{1^{k}}{(k+1)!} = \sum_{k=0}^{\infty} \frac{1}{(k+1)!}\). This shows that the value of the infinite sum \(\sum_{k=0}^{\infty} \frac{1}{(k+1)!}\) is equal to \(f(1)\), which can be evaluated using the original function definition.

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주요 개념

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Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives at a single point. For a function f(x) centered at 0, it is expressed as f(x) = Σ (f⁽ᵏ⁾(0)/k!) xᵏ. This allows approximation and evaluation of functions like f(x) = (eˣ − 1)/x near zero.

Maclaurin Series for the Exponential Function

The Maclaurin series for eˣ is Σ (xᵏ / k!) from k=0 to ∞. Using this, eˣ − 1 can be written as Σ (xᵏ / k!) for k=1 to ∞. Dividing by x shifts the index, which helps express f(x) as a power series and evaluate it at specific points like x=1.

Infinite Series Summation and Evaluation

Evaluating infinite series involves recognizing patterns and using known series sums. Here, the series ∑ₖ₌₀∞ 1/(k+1)! corresponds to the sum of reciprocals of factorials shifted by one index, which relates to the exponential function minus 1. Understanding this helps find exact values of such sums.