Partial Fraction Decomposition and Sums (Difference Sums)
When decomposed into a difference of consecutive terms, such as 1/{k(k+1)} = 1/k − 1/(k+1), the terms successively cancel each other (a telescoping sum).
Example ① Find Σ_{k=1}^{n} 1/{k(k+1)}.
Solution: Each term = 1/k − 1/(k+1).
Sum = (1 − 1/2) + (1/2 − 1/3) + … + (1/n − 1/(n+1)) = 1 − 1/(n+1) = n/(n+1)
Solution: Each term = 1/k − 1/(k+1).
Sum = (1 − 1/2) + (1/2 − 1/3) + … + (1/n − 1/(n+1)) = 1 − 1/(n+1) = n/(n+1)
Key Point
- Product of two consecutive terms in the denominator → decompose into a difference
- Can also be applied to forms such as 1/{k(k+1)(k+2)}
Practice Problems
- Σ_{k=1}^{n} 1/{(2k−1)(2k+1)}
Answer and Explanation
- Answer:n/(2n+1)
1/{(2k−1)(2k+1)} = (1/2){1/(2k−1) − 1/(2k+1)}.