Problem 51 - Algebra and Trigonometry

Question

Find the partial fraction decomposition for \(\frac{1}{x(x+1)}\) and use the result to find the following sum: $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{99 \cdot 100}$$

Step-by-Step Solution

Answer

The sum of the series is \( \frac{99}{100} \).

Step 1: Find the partial fraction decomposition

We can express the given fraction, \( \frac{1}{x(x+1)} \), as the sum of two simpler fractions: \( \frac{A}{x} \) and \( \frac{B}{x+1} \), for some coefficients A and B: \n\n\( \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \)\n\nMultiplying throughout by the common denominator, \( x(x+1) \), gives us: \n\n\( 1 = A(x+1) + Bx \)\n\nBy comparing coefficients, we can determine A and B. Setting x = 0 gives A = 1, and setting x = -1 gives B = -1. So, our partial fraction decomposition is: \n\n\( \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1} \)

Step 2: Express the given sum using partial fractions

Using the partial fraction decomposition we've found, we can express each term of the given sum, \( \frac{1}{n(n+1)} \), as the difference between two simpler fractions: \n\n\( \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \)\n\nSo, the given sum can be written as: \n\n\( \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+ \dots+\frac{1}{99 \cdot 100} = (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}- \frac{1}{4})+ \dots+(\frac{1}{99}-\frac{1}{100}) \)

Step 3: Calculate the sum

Now observe that in the sum obtained in the previous step, every term from \( \frac{1}{2} \) to \( \frac{1}{99} \) appears twice and gets cancelled out. We only have the first and the last term remaining: \n\n\( 1 - \frac{1}{100} \)\n\nSo, the sum of the given series is \( \frac{99}{100} \).