Problem 42 - Calculus

Question

By recognizing each series as a Taylor series evaluated at a particular value of \(x,\) find the sum of each of the following convergent series. $$1-\frac{1}{2 !}+\frac{1}{4 !}-\frac{1}{6 !}+\cdots$$

Step-by-Step Solution

Answer

The sum is \(\cos(1)\).

Step 1: Identify the Series Pattern

The given series is \(1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \cdots\). This series is alternating, with terms following the pattern \((-1)^n \frac{1}{(2n)!}\) where \(n\) starts from 0. The series includes only the even factorials in the denominators.

Step 2: Relate the Series to a Known Taylor Series

The Taylor series for \(\cos(x)\) is given by: \[\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\]Comparing this with our series pattern, we see that the series is missing the variable \(x\) raised to even powers. This implies \(x = 1\) in \(\cos(x)\).

Step 3: Evaluate the Cosine Function

Substitute \(x = 1\) into the Taylor series for \(\cos(x)\):\[\cos(1) = \sum_{n=0}^{\infty} \frac{(-1)^n 1^{2n}}{(2n)!} = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \cdots\]Therefore, the sum of the series is equal to \(\cos(1)\).

Step 4: Provide Final Answer

The sum of the given convergent series is \(\cos(1)\), by recognizing it as a Taylor series for the cosine function evaluated at \(x = 1\).

Key Concepts

Convergent Series

A convergent series is a series whose terms decrease and approach a finite limit. In other words, the sum of its terms adds up to a specific, finite number. This concept is essential when analyzing series such as Taylor series used in mathematics.
Taylor series are special kinds of convergent series where the sum of an infinite sequence of terms is used to represent functions. For these series to be useful, they must converge to the actual value of the function they represent within an interval of convergence.
With respect to the exercise at hand, the series given \(1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \cdots\) converges to a finite value, specifically to the result of \(\cos(1)\) when evaluated as part of the cosine function's Taylor series.

• The series converges since its terms get smaller and ultimately sum to a finite number, \(\cos(1)\).

Cosine Function

The cosine function, often represented as \(\cos(x)\), is a fundamental trigonometric function. It's periodic and oscillates between -1 and 1. One remarkable aspect of this function is its representation using an infinite series known as a Taylor series:\[\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\]This series expresses \(\cos(x)\) as a sum of terms involving even powers of \(x)\) and alternating signs.
This mathematical representation is precise and simplifies calculations in many areas, including physics and engineering. In our exercise, the given series matches the form of the Taylor series for the cosine function when \(x = 1)\).

• This connection allows us to equate the sum of the series directly to \(\cos(1)\).

Alternating Series

An alternating series is characterized by the alternating sign of its terms. This means the series has a pattern of positive and negative terms. In mathematics, alternating series are especially significant because they can still converge even if infinitely many terms are involved.The series provided in the exercise,\(1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \cdots\), exemplifies an alternating series:

• Its terms alternate between positive and negative.
• The pattern is dictated by \((-1)^n\), ensuring that each term alternates signs with successive terms.

This specific type of series is crucial in the convergence test known as the Alternating Series Test, which helps determine if the series converges to a finite limit. Here, this alternating behavior aligns perfectly with the structure of the Taylor series for \(\cos(x)\), which relies on alternating terms.