\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}
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- = 16 - 40 + 24
- Question:** A mathematician is analyzing a topological space and needs to compute the sum of the series \( \sum_{k=1}^{50} \frac{1}{k(k+1)} \) as part of a homology calculation. Calculate this sum.
- Solution:** The given series is \( \sum_{k=1}^{50} \frac{1}{k(k+1)} \). We can simplify the terms using partial fraction decomposition:
- Solving \( 1 = A(k+1) + Bk \), we equate coefficients:
- \( A + B = 0 \)
- \( A = 1 \)
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