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.
["Title: Efficiently Computing the Series ( \sum_{k=1}^{50} \frac{1}{k(k+1)} ) for Topological Homology Calculations", "In algebraic topology, series summation often plays a crucial role—especially when determining invariants through chain complexes. For mathematicians working with homology groups, simplifying series like ( \sum_{k=1}^{50} \frac{1}{k(k+1)} ) is essential for both conceptual clarity and computational efficiency.", "### The Series in Question\nThe target sum is:", "[\n\sum_{k=1}^{50} \frac{1}{k(k+1)}\n]", "This is a classic telescoping series, widely used in mathematical analysis due to its elegant closed-form behavior.", "### Why This Series Matters in Topology\nSuch series frequently appear when evaluating generating functions or when computing boundary rulings in simplicial complexes. Simplifying this sum allows mathematicians to redirect attention toward structural properties rather than brute-force computation—critical in high-dimensional homology theory.", "### Mathematical Simplification via Partial Fractions\nThe term ( \frac{1}{k(k+1)} ) can be decomposed using partial fraction decomposition:", "[\n\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}\n]", "Multiplying both sides by ( k(k+1) ) gives:\n( 1 = A(k+1) + Bk )", "Expanding and equating coefficients:\n- For ( k^1 ): ( A + B = 0 )\n- For ( k^0 ): ( A = 1 )\nThus, ( A = 1 ), ( B = -1 ), so:", "[\n\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\n]", "### Transforming the Sum into a Telescoping Series\nSubstituting the simplified form into the original sum:", "[\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+1} \right)\n]", "Writing the first few and last few terms:", "[\n\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{50} - \frac{1}{51} \right)\n]", "Observe that intermediate terms cancel:\n( -\frac{1}{2} + \frac{1}{2}, -\frac{1}{3} + \frac{1}{3}, \ldots, -\frac{1}{50} + \frac{1}{50} )", "Only the first and last terms survive:", "[\n\frac{1}{1} - \frac{1}{51}\n]", "### Final Computation\n[\n\sum_{k=1}^{50} \frac{1}{k(k+1)} = 1 - \frac{1}{51} = \frac{51 - 1}{51} = \frac{50}{51}\n]", "### Conclusion\nBy recognizing the telescoping nature of the series and applying partial fraction decomposition, the computation reduces cleanly to ( \frac{50}{51} ). This result exemplifies how foundational algebraic techniques streamline complex calculations in topological homology, enabling mathematicians to focus on deeper structural insights.", "Whether used in chain complex evaluations or generating function manipulations, mastering such sums strengthens both theoretical rigor and computational fluency in modern topology.", "---", "Keywords:\nseries summation, telescoping series, topological homology, ( \sum_{k=1}^{n} \frac{1}{k(k+1)} ), partial fractions, algebraic topology, homology calculations, mathematical series simplification.\nMeta Description:\nEfficiently compute ( \sum_{k=1}^{50} \frac{1}{k(k+1)} ) using telescoping series techniques—essential for simplifying homology-related sums in algebraic topology."]
