The series \(\sum_{n=1}^{50} \frac{1}{n(n+1)}\) is a telescoping series. We simplify each term using partial fractions:
["Understanding the Telescoping Series: (\sum_{n=1}^{50} \frac{1}{n(n+1)})", "When studying infinite and finite summation, few techniques are as elegant and powerful as recognizing a telescoping series. One of the most famous examples is the finite sum:", "[\n\sum_{n=1}^{50} \frac{1}{n(n+1)}\n]", "This sum beautifully simplifies using partial fractions, revealing a pattern where most terms cancel, leaving only the first and last components. In this article, we explore why this series is telescoping and demonstrate its full simplification step-by-step.", "---", "### What Is a Telescoping Series?", "A telescoping series is a sum in which many intermediate terms cancel out, reducing the entire sum to a few simple expressions. This cancellation happens because each term can be written as the difference of two adjacent values—much like terms in a telescoping thermometer section collapsing step by step.", "---", "### Why Is (\sum_{n=1}^{50} \frac{1}{n(n+1)}) Telescoping?", "The key to this telescoping behavior lies in decomposing (\frac{1}{n(n+1)}) using partial fraction decomposition. By expressing the term as a difference of simple fractions, we expose the structure that allows cancellation across the sequence.", "---", "### Step 1: Decompose Using Partial Fractions", "We begin by writing:", "[\n\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}\n]", "Multiply both sides by (n(n+1)):", "[\n1 = A(n+1) + Bn\n]", "Expanding and combining like terms:", "[\n1 = An + A + Bn = (A + B)n + A\n]", "For this identity to hold for all (n), the coefficients must satisfy:", "[\n\begin{cases}\nA + B = 0 \\nA = 1\n\end{cases}\n]", "Solving this system, we get (A = 1) and (B = -1). Thus:", "[\n\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\n]", "---", "### Step 2: Rewrite the Sum", "Substitute the decomposition into the original sum:", "[\n\sum_{n=1}^{50} \frac{1}{n(n+1)} = \sum_{n=1}^{50} \left( \frac{1}{n} - \frac{1}{n+1} \right)\n]", "Now write out the terms explicitly:", "[\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]", "---", "### Step 3: Observe the Cancellation (Telescoping Effect)", "Notice that the (-\frac{1}{2}) from the first term cancels with the (\frac{1}{2}) from the second. Similarly, (-\frac{1}{3}) cancels with (+\frac{1}{3}), and so on. This cancellation continues all the way through the sequence.", "Only two terms remain uncanceled:", "- The positive first term: (\frac{1}{1} = 1)\n- The negative last term: (-\frac{1}{51})", "Thus, the entire sum simplifies elegantly to:", "[\n\sum_{n=1}^{50} \frac{1}{n(n+1)} = 1 - \frac{1}{51}\n]", "---", "### Step 4: Compute the Final Value", "Simplify the expression:", "[\n1 - \frac{1}{51} = \frac{51}{51} - \frac{1}{51} = \frac{50}{51}\n]", "---", "### Conclusion", "The series (\sum_{n=1}^{50} \frac{1}{n(n+1)}) is a paradigmatic example of a telescoping series. By applying partial fractions, we revealed a cancellation pattern where intermediate terms vanish, leaving only boundary contributions. The final value is:", "[\n\boxed{\frac{50}{51}}\n]", "Such techniques not only simplify computations but also deepen insight into the structure of infinite and finite sums. Recognizing telescoping series saves time and reveals mathematical beauty in seemingly complex sums.", "---", "Keywords: telescoping series, partial fractions, summation, (\sum_{n=1}^{50} \frac{1}{n(n+1)}), simplification, telescoping effect, mathematical technique, infinite series simplification."]
