91Ó°ÊÓ

Partial fraction decomposition is a technique often used to simplify complex rational expressions and integrands. It involves breaking down a fraction with a polynomial in the denominator into simpler fractions that are easier to integrate or sum in the context of series. In the case of the infinite series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\), the decomposition is particularly straightforward.

Consider the general form of a fraction where the denominator can be factored into linear factors, \(\frac{1}{(ax+b)(cx+d)}\). To decompose, we would find constants \(A\) and \(B\) such that \(\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}\). Finding these constants typically involves equating coefficients or using a system of equations to solve for \(A\) and \(B\).

Applying this process to \(\frac{1}{n(n+1)}\), we achieve a simple subtraction between two fractions: \(\frac{1}{n} - \frac{1}{n+1}\). This step is crucial as it sets us up for the process that leads to simplification and eventually convergence, known as telescoping.

A telescoping series is a particular type of series where each term cancels out part of the next term. This cancellation leads to a significant simplification when trying to evaluate the sum of the series. For the series in question, \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\), after applying partial fraction decomposition, we notice the structure \(\frac{1}{n} - \frac{1}{n+1}\) leads to a pattern where each negative fraction nullifies the positive fraction of the preceding term.

The beautiful aspect of a telescoping series is how it collapses. Imagine each term as a segment of a retractable telescope, with successive segments collapsing into each other, leaving only the first and last segments visible. In our series, after writing out a few terms, it becomes clear that all intermediate terms will vanish, and we are left with just the first and last term to consider for the evaluation of the series.

Series evaluation can sometimes be complex and rely on sophisticated mathematical tools. However, with telescoping series like the one in our exercise, the evaluation becomes much more manageable. After the collapsing of terms, we only need to consider the 'first' and 'last' term of our series, recognizing that intermediate terms cancel each other out.

In the case of the given series \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\), after employing partial fraction decomposition and identifying it as a telescoping series, we're left with the terms \(1 - \frac{1}{n+1}\) as \(n\) approaches infinity. By taking the limit as \(n\) goes to infinity, we note that \(\frac{1}{n+1}\) approaches zero, and thus the sum of the series is simply \(1 - 0 = 1\). This relatively simple process illustrates how identifying the nature of a series can greatly simplify the task of evaluating it.