Problem 8 - Calculus: One and Several Variables

Question

Evaluate. $$\sum_{k=1}^{5} \frac{(-1)^{2}}{k !}$$

Step-by-Step Solution

Answer

The sum of the series \(\sum_{k=1}^{5} \frac{(-1)^{2}}{k!}\) evaluates to approximately 1.72.

Step 1: Understanding the general term

The given summation is \(\sum_{k=1}^{5} \frac{(-1)^{2}}{k!}\), where the general term is \(\frac{(-1)^2}{k!}\). Since \((-1)^2 = 1\), the general term simplifies to \(\frac{1}{k!}\). Now, we will evaluate this expression for each value of k from 1 to 5.

Step 2: Evaluating the sum

We can now substitute the values k = 1, 2, 3, 4, and 5 into the general term \(\frac{1}{k!}\) and add them up to get the sum: \(S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}\) First, we will evaluate each factorial and then find the sum.

Step 3: Evaluating the factorials

Now, let's compute the factorials for each term in the sum: \(1! = 1\) \(2! = 1 \cdot 2 = 2\) \(3! = 1 \cdot 2 \cdot 3 = 6\) \(4! = 1 \cdot 2 \cdot 3 \cdot 4 = 24\) \(5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120\) Now, let's substitute these factorials into the sum.

Step 4: Computing the sum

Finally, we will substitute the factorials we calculated back into the sum S: \(S = \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120}\) Now, we can add up the terms: \(S = 1 + 0.5 + 0.166666... + 0.041666... + 0.008333...\) Rounding each term to 2 decimal places and adding them up: \(S \approx 1 + 0.5 + 0.17 + 0.04 + 0.008\) \(S \approx 1.72\) So, the evaluated sum is approximately 1.72.

Key Concepts

Factorials

Factorials are an essential concept in calculus and many areas of mathematics. A factorial, denoted by the exclamation mark (!), refers to the product of all positive integers up to a certain number. For instance, the factorial of a number \( n \) is represented as \( n! \), which is the multiplication of all whole numbers from 1 to \( n \).

• For example, \( 3! = 1 \cdot 2 \cdot 3 = 6 \).
• As another example, \( 5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 \).

The factorial function grows very quickly with larger numbers. It is important for calculating combinations and permutations, and plays a key role in many mathematical series and equations. In this exercise, each term of the series involved calculating a factorial, which was then used to determine the value of the general term. For example, for \( k = 3 \), the corresponding factorial is \( 3! = 6 \), affecting the calculation of the term \( \frac{1}{3!} \).

Summation

Summation, often denoted by the Greek letter sigma (\( \Sigma \)), represents the addition of a sequence of numbers. It is a fundamental concept in calculus, providing an effective way to express series and calculate their totals. A series is simply the sum of the terms of a sequence.In the exercise given, the concept of summation is applied to find the sum of a finite series, represented as:\[ \sum_{k=1}^{5} \frac{1}{k!} \]Here's how it unfolds:

• Start from \( k = 1 \), substitute into the general term \( \frac{1}{k!} \), and calculate the value.
• Repeat for each integer up to 5.
• Sum all the individual values obtained to find the final sum.

The result of this particular summation gives an approximation of 1.72 when all terms are added together, showcasing how summation helps in accumulating the series elements.

General Term

The general term of a series is crucial to its definition and subsequent calculations, as it defines each term of the sequence. In a series, the general term is often represented by an expression that includes a variable, typically \( k \) or \( n \), standing for the position number in the series.For the series given in the exercise, the general term was initially expressed as \( \frac{(-1)^{2}}{k!} \). However, since \((-1)^2 = 1\), this simplifies to \( \frac{1}{k!} \). This simplification is a vital step as it clarifies the actual contributions of each term when calculating the summation.

• The function of the general term is to provide the formula that will repeatedly calculate values for each term in the sequence.
• Understanding the general term ensures that each element of the series is correctly identified and added to the sum.

For example, plug in \( k = 3 \) to calculate the third term: \( \frac{1}{3!} = \frac{1}{6} \), and repeat similarly for all others. Mastery of determining and manipulating the general term is fundamental in studying and evaluating series.