Evaluating a Complex Series: A Step-by-Step Guide

This article provides a detailed explanation of how to evaluate the series

[sum_{k 1}^n frac{k - 3}{k(k 1)(k 2)}]

Introduction to the Series

Mathematical series are a fundamental concept in mathematics, with applications across various fields such as physics, engineering, and computer science. This article focuses on a specific type of series:

[sum_{k 1}^n frac{k - 3}{k(k 1)(k 2)}]

Step-by-Step Solution

Step 1: Identifying the nth Term

The first step is to identify the nth term of the series, which in this case is:

[frac{k - 3}{k(k 1)(k 2)}]

Step 2: Decomposing the nth Term into Fractions

Decomposing the nth term into a sum of simpler fractions can often simplify the process of summing the series. Let's break down the term into:

[frac{x}{k} - frac{y}{k 1} frac{z}{k 2}]

Our goal is to find the values of x, y, and z. To do this, we set up the equation:

[frac{k - 3}{k(k 1)(k 2)} frac{x}{k} - frac{y}{k 1} frac{z}{k 2}]

By multiplying each side by the common denominator (k(k 1)(k 2)), we get:

[k - 3 x(k 1)(k 2) - yk(k 2) zk(k 1)]

Expanding this equation gives:

[k - 3 x(k^2 3k 2) - yk^2 - 2yk zk^2 zk]

Simplifying the right-hand side:

[k - 3 (x - y z)k^2 (3x - 2y z)k (2x)]

By comparing coefficients, we get the following system of equations:

[x - y z 0] [3x - 2y z 1] [2x -3]

Step 3: Solving for x, y, and z

Solving the system of equations:

From [2x -3], we get [x -frac{3}{2}] Substituting [x -frac{3}{2}] into [x - y z 0], we get [-frac{3}{2} - y z 0], which simplifies to [y - z -frac{3}{2}] Substituting [x -frac{3}{2}] into [3x - 2y z 1], we get [-frac{9}{2} - 2y z 1], which simplifies to [-2y z frac{11}{2}]

Solving these two equations:

From [y - z -frac{3}{2}], we have [y z - frac{3}{2}] Substituting into [-2y z frac{11}{2}], we get [-2(z - frac{3}{2}) z frac{11}{2}], which simplifies to [-2z 3 z frac{11}{2}], or [-z frac{5}{2}], hence [z -frac{5}{2}] Substituting [z -frac{5}{2}] back, we get [y -frac{5}{2} - frac{3}{2} -4]

Step 4: Evaluating the Series

Now we have decomposed the term into:

[frac{k - 3}{k(k 1)(k 2)} -frac{3}{2k} frac{4}{k 1} - frac{5}{2(k 2)}]

When we sum this series, many terms will cancel out. Let's write out the sum for the first few terms:

[left(-frac{3}{2 cdot 1} frac{4}{2} - frac{5}{2 cdot 3}right) left(frac{4}{2} - frac{5}{2 cdot 3} - frac{3}{2 cdot 2} frac{4}{3} - frac{5}{4}right) cdots left(frac{4}{n} - frac{5}{2(n 1)} - frac{3}{2n} frac{5}{2(n 2)}right)]

This results in many cancellation terms, leading to:

[left(-frac{3}{2 cdot 1} - frac{3}{2n}right) left(frac{4}{2} - frac{4}{n 1}right) - left(frac{5}{2 cdot 3} - frac{5}{2(n 2)}right)]

Simplifying this, we get:

[left(-frac{3}{2} - frac{3}{2n}right) left(2 - frac{4}{n 1}right) - left(frac{5}{6} - frac{5}{2(n 2)}right)]

Combining all terms, we get:

[left(-frac{3}{2} - frac{3}{2n}right) left(2 - frac{4}{n 1}right) - left(frac{5}{6} - frac{5}{2(n 2)}right)]

For large (n), the terms (frac{3}{2n}), (frac{4}{n 1}), and (frac{5}{2(n 2)}) approach zero, so we have:

[left(-frac{3}{2}right) 2 - frac{5}{6} frac{-9 12 - 5}{6} frac{4}{6} frac{2}{3}]

Thus, the final result is:

[sum_{k 1}^n frac{k - 3}{k(k 1)(k 2)} frac{5}{4} - frac{1}{n(n 1)}]

For large (n), this simplifies to:

[sum_{k 1}^infty frac{k - 3}{k(k 1)(k 2)} frac{5}{4}]

Conclusion

By understanding the process of decomposing fractions and identifying common patterns, the evaluation of complex series can be simplified. This technique is not only useful for this specific series, but it can also be applied to other series with similar structures.