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

To solve the problem of finding the value of the series \( \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \ldots + \frac{99}{100!} \), we can recognize a pattern in how each term is constructed.

Each term in the series can be expressed as follows:

\(\frac{k}{(k+1)!} = \frac{k+1-1}{(k+1)!} = \frac{k+1}{(k+1)!} - \frac{1}{(k+1)!}\)

This identity simplifies to:

\(\frac{k}{(k+1)!} = \frac{1}{k!} - \frac{1}{(k+1)!}\)

Using this transformation, the series becomes a telescoping series:

\(\left(\frac{1}{1!} - \frac{1}{2!}\right) + \left(\frac{1}{2!} - \frac{1}{3!}\right) + \dots + \left(\frac{1}{99!} - \frac{1}{100!}\right)\)

Observe that all intermediate terms cancel each other out, leaving:

\(\frac{1}{1!} - \frac{1}{100!}\)

Therefore, the sum of the series is:

\(1 - \frac{1}{100!}\)

To express this sum in the form of a single fraction:

\(1 - \frac{1}{100!} = \frac{100!}{100!} - \frac{1}{100!} = \frac{100! - 1}{100!}\)

Thus, the value of the series is \( \frac{100! - 1}{100!} \).

The correct answer is therefore:

\( \frac{100! - 1}{100!} \)

Answer 2

The given series is:

\(\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{99}{100!}\)

We can denote this sum by \( S \):

\( S = \sum_{n=2}^{100} \frac{n-1}{n!} \)

Rewriting each term:

\(\frac{n-1}{n!} = \frac{n}{n!} - \frac{1}{n!} = \frac{1}{(n-1)!} - \frac{1}{n!}\)

Substitute this expression back into the sum:

\( S = \sum_{n=2}^{100} \left( \frac{1}{(n-1)!} - \frac{1}{n!} \right) \)

This is a telescoping series. Most terms will cancel out, leaving:

\( S = 1 - \frac{1}{100!} \)

By simplifying, we have:

\( S = \frac{100!}{100!} - \frac{1}{100!} = \frac{100! - 1}{100!} \)

Thus, the value of the series is \(\frac{100! - 1}{100!}\)