Problem 41 - Algebra and Trigonometry

Question

find each indicated sum. $$ \sum_{i=1}^{5} \frac{i !}{(i-1) !} $$

Step-by-Step Solution

Answer

The sum of the series is 15.

Step 1: Understand the factorial

The factorial of a number \(n\) is the product of all positive integers less than or equal to \(n\). Therefore, the factorial of \(i\) is symbolized as \(i!\) and it's equal to \(i * (i-1) * (i-2) * ... * 3 * 2 * 1\). The factorial of \(i-1\) is therefore equal to \((i-1) * (i-2) * ... * 3 * 2 * 1\). When \(i!\) is divided by \((i-1)!\), all terms cancel out except for \(i\), because \(\frac{i * (i-1) * (i-2) * ... * 3 * 2 * 1}{(i-1) * (i-2) * ... * 3 * 2 * 1} = i\).

Step 2: Apply understanding to the series

Now we can apply this understanding to the series. We're asked to find the sum of \(\frac{i !}{(i-1) !}\) from \(i=1\) to \(5\). Given our understanding from Step 1, we know that this is equal to \(i\). Therefore, the sum we're looking for is \(1+2+3+4+5\).

Step 3: Calculate the sum

Now we can simply calculate the sum of integers from \(1\) to \(5\). This sum is equal to \(1+2+3+4+5=15\).

Key Concepts

Summation Notation

Summation notation, often represented by the sigma symbol \(\Sigma\), is used to denote the sum of a sequence of numbers. It's a concise and efficient way to express the addition of a series of numbers without explicitly writing them all out. In the notation \(\sum_{i=1}^{n} a_i\), the symbol on the bottom, \(i=1\), is called the index of summation and indicates the starting value of \(i\). The symbol \(n\) on top of \(\Sigma\) represents the endpoint of the summation.

Here is how to understand and use summation notation:

• Identify the series to be summed and recognize the variable part, which is represented by \(i\) in our example.
• Determine the initial and final values of the index. For instance, from the problem, \(i\) starts at 1 and ends at 5.
• Formulate the term to sum as a function of \(i\), such as \(\frac{i!}{(i-1)!}\).

The notation is powerful, especially when dealing with large series, because it synthesizes the operation into a singular expression.

Series Evaluation

Evaluating a series involves determining the sum of terms generated by an expression within a defined range as specified by summation notation. In our exercise, the task is to evaluate the series \(\sum_{i=1}^{5} \frac{i!}{(i-1)!}\).

Let's break down the steps needed for evaluating this series:

• First, simplify the expression inside the summation. In this problem, \(\frac{i!}{(i-1)!}\) can be simplified to \(i\). This is because factorial division cancels out a sequence of successive numbers, revealing just the factor \(i\).
• Substitute this simplified form back into the summation: \(\sum_{i=1}^{5} i\). This means we simply add up the numbers from 1 to 5.
• Compute the sum: \(1+2+3+4+5\), which totals up to 15.

With these steps, evaluating the series becomes a manageable process. It simplifies a potentially complex problem into an easily computable sum of integers.

Integer Summation

Integer summation is foundational in mathematics and is simply the process of adding whole numbers together to find their total. It’s a straightforward task but forms the basis for more intricate mathematical computations and series evaluations.

In the context of our exercise:

• Recognize that when the simplified form of a series consists of integers only, the task is reduced to summing these integers over the given range.
• Using the exercise example, the series was reduced to \(1+2+3+4+5\).
• Perform the addition step-by-step, often using short cuts like pairing numbers if a long series, or applying formulas for consecutive sums. However, for series up to 5, direct addition suffices, resulting in sum 15.

Understanding how to efficiently sum integers ensures you can tackle more complex series and sequences in mathematical studies.