Question

Find the value of $(1 - \frac{1}{2})(1 - \frac{1}{3})(1 - \frac{1}{4}) \dots (1 - \frac{1}{n})$.

• A. $\frac{1}{n}$
• B. $\frac{1}{n+1}$
• C. $1 + \frac{1}{n}$
• D. $1 - \frac{1}{n}$

Hint:

For products or sums of series, always simplify the first few terms to identify a telescoping pattern where intermediate terms cancel out. This is a common technique for solving such problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks to find the value of a product series involving fractions.

Step 2: Key Formula or Approach:
Simplify each term in the product, then look for a pattern of cancellation (telescoping product).

Step 3: Detailed Explanation:
Given series:
\[ \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \dots \left(1 - \frac{1}{n}\right) \]
Simplify each term:
\[ 1 - \frac{1}{2} = \frac{2-1}{2} = \frac{1}{2} \]
\[ 1 - \frac{1}{3} = \frac{3-1}{3} = \frac{2}{3} \]
\[ 1 - \frac{1}{4} = \frac{4-1}{4} = \frac{3}{4} \]
And so on, up to the last term:
\[ 1 - \frac{1}{n} = \frac{n-1}{n} \]
Now, write the product:
\[ \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{3}{4}\right) \times \dots \times \left(\frac{n-1}{n}\right) \]
Observe the cancellation pattern: the numerator of each term cancels with the denominator of the preceding term.
\[ \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{\cancel{4}} \times \dots \times \frac{\cancel{n-1}}{n} \]
Only the first numerator (1) and the last denominator (n) remain.
The value of the product is $\frac{1}{n}$.

Step 4: Final Answer:
The value of the series is $\frac{1}{n}$.