Question

The value of $ \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ... + \frac{99}{100!} $ is equal to

• A. \( \frac{100! - 1}{100!} \)
• B. \( \frac{100! + 1}{100!} \)
• C. \( \frac{999! - 1}{999!} \)
• D. \( \frac{999! + 1}{999!} \)

Hint:

Factorial-based series often simplify using telescoping series methods or recognizing relationships between terms that simplify the expression.

Answer 1

The Correct Option is A

The given series is: \[ S = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \dots + \frac{99}{100!} \] This is a series with terms of the form \( \frac{n}{(n+1)!} \).
By analyzing the structure of the series and summing the individual terms using properties of factorials, we arrive at the final result, which simplifies to: \[ S = \frac{100! - 1}{100!} \]