91Ó°ÊÓ

Partial Fraction Decomposition is a technique used to break down a complex rational expression into simpler fractions, which are easier to work with. In this exercise, we aim to decompose the fraction \( \frac{1}{n^2-1} \) into two simpler fractions. This process often helps us identify patterns and make the series easier to sum.

The denominator \( n^2 - 1 \) can be factored into \((n-1)(n+1)\). This factorization allows us to express the original fraction as a sum of two fractions with linear denominators: \( \frac{A}{n-1} + \frac{B}{n+1}\).

To find the values of \( A \) and \( B \), we equate and solve: \ 1 = A(n+1) + B(n-1).
We then substitute test values for \( n \). By setting \( n = 1 \), the equation simplifies to find \( A = \frac{1}{2} \). Similarly, setting \( n = -1 \), we find \( B = -\frac{1}{2} \).

The original fraction is thus decomposed into \( \frac{1}{2(n-1)} - \frac{1}{2(n+1)} \). This decomposition is particularly helpful in simplifying and summing the series.

A telescoping series is a series where many terms cancel out, making it much simpler to find the sum. After applying partial fraction decomposition to our series, it takes the form \( \sum_{n=2}^{\infty} \left( \frac{1}{2(n-1)} - \frac{1}{2(n+1)} \right) \).

Writing out the first few terms demonstrates the telescoping nature:

• For \( n=2 \), the terms are \( \frac{1}{2}(\frac{1}{1} - \frac{1}{3}) \)
• For \( n=3 \), we have \( \frac{1}{2}(\frac{1}{2} - \frac{1}{4}) \)
• For \( n=4 \), \( \frac{1}{2}(\frac{1}{3} - \frac{1}{5}) \) etc.

As observed, consecutive terms \( \frac{1}{k} \) and \( -\frac{1}{k} \) from adjacent fractions cancel each other out. This cancellation happens infinitely within the series. The ultimate result is a reduced series that makes arriving at a final sum straightforward. In our case, significant cancellation occurs, leaving few terms to consider when evaluating the total sum.

Calculating the sum of a telescoping series involves focusing on the non-cancelled terms that remain after all possible cancellations. For this exercise, we identified a converging series through decomposition and term cancellation.

Recall that much of our series cancels out, leaving only a few initial terms. Observing the expanded form, we notice the remaining terms after cancellation are \( \frac{1}{2}(\frac{1}{1}) \) and \( \frac{1}{2}(\frac{1}{2}) \).

The sum of these remaining terms is straightforward:

• \( \frac{1}{2}(1) = \frac{1}{2} \)
• \( \frac{1}{2}(\frac{1}{2}) = \frac{1}{4} \)

Adding these gives the total sum of the series: \( \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \).

Thus, the sum of the original convergent series \( \sum_{n=2}^{\infty} \frac{1}{n^2-1} \) is \( \frac{3}{4} \), providing a clear solution through the power of sequence decomposition and term cancellation.