Chapter 12: Problem 8
Evaluate. $$\sum_{k=1}^{5} \frac{(-1)^{2}}{k !}$$
Short Answer
Expert verified
The sum of the series \(\sum_{k=1}^{5} \frac{(-1)^{2}}{k!}\) evaluates to approximately 1.72.
Step by step solution
01
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.
02
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.
03
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.
04
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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, \( 5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120 \).
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.
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.