Problem 5 - A Gentle Introduction to the Art of Mathematics

Question

Prove the following formula for a product. $$ \prod_{i=2}^{n}\left(1-\frac{1}{i}\right)=\frac{1}{n} $$

Step-by-Step Solution

Answer

\( \prod_{i=2}^{n}\left(1-\frac{1}{i}\right)=\frac{1}{n} \)

Step 1 - Write the Product Formula as a Series of Terms

Start by writing out the product formula as a series of terms: i=2 to n: \textrm{Product} = \prod_{i=2}^{n}\left(1-\frac{1}{i}\right)This gives: \( \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{n}\right) \)

Step 2 - Simplify Each Term Individually

Simplify each term individually:\( \left(1 - \frac{1}{i}\right) = \frac{i-1}{i} \)So the product becomes:\( \prod_{i=2}^{n} \left(1 - \frac{1}{i}\right) = \prod_{i=2}^{n} \frac{i-1}{i} \)

Step 3 - Rewrite the Product

Rewrite the product in a more manageable form:\( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n} \)

Step 4 - Cancel Out Terms

This can be simplified by noticing that many terms cancel out:\( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n} = \frac{1 \cdot 2 \cdot 3 \cdots (n-1)}{2 \cdot 3 \cdot 4 \cdots n} \)Most of the terms in the numerator and denominator cancel out, leaving only:\( \frac{1}{n} \)

Step 5 - Conclude

The product can be written as:\( \prod_{i=2}^{n}\left(1-\frac{1}{i}\right)=\frac{1}{n} \)This completes the proof.

Key Concepts

product formula

A product formula is used to describe a series of multiplicative terms.
It's typically expressed using the product notation, represented by the capital Greek letter pi ( \prod ).
Applying this concept, we want to prove: \[ \prod\_{i=2}\^{n}\left(1-\frac{1}{i}\right)=\frac{1}{n}\] Here’s how we approach it step-by-step:
First, expand the product formula into individual terms: \[ \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{n}\right)\]

simplification techniques

Simplification techniques help make complex expressions easier to manage.
By dealing with each term individually, we simplify our original product: \[ \left(1 - \frac{1}{i}\right) = \frac{i-1}{i}\] Using this, our product becomes:
\[ \prod\_{i=2}\^{n}\left(1 - \frac{1}{i}\right) = \prod\_{i=2}\^{n} \frac{i-1}{i}\]
Then, rewrite it as: \[ \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n}\]
Here, we notice that most terms in the numerator and denominator cancel out: \[ \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n} = \frac{1 \cdot 2 \cdot 3 \cdots (n-1)}{2 \cdot 3 \cdot 4 \cdots n}\] Resulting in:
\[ \frac{1}{n}\]

series and sequences

In mathematics, a series is the sum of the terms of a sequence.
A sequence is an ordered list of numbers.
Here, we are dealing with a product series.
To understand this better, consider:
A sequence of terms like \left(a_1, a_2, a_3, \ldots, a_n\right).
If you multiply these terms together in a product series, you get a product formula.
Using our example:
\[ b_i = \left(1 - \frac{1}{i}\right)\]
As shown, each term simplifies down and we get a product that proves our initial formula.
This process illustrates the power of sequences and series in simplifying and solving mathematical problems, showing that understanding how terms cancel each other is key to achieving the final result: \[ \prod\_{i=2}\^{n}\left(1-\frac{1}{i}\right)=\frac{1}{n}\]