Question

\(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{2}-n}\)

Short answer

The sum of the series is 1.

Step-by-step solution

Identify the Series

The given series is \(\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n^{2}-n}\).

Simplify the Denominator

Rewrite the denominator \(n^{2} - n\) as \((n)(n-1)\). Thus, the term becomes \(\frac{(-1)^{n}}{n(n-1)}\).

Decompose into Partial Fractions

Express \(\frac{1}{n(n-1)}\) using partial fractions: \(\frac{1}{n(n-1)} = \frac{A}{n} + \frac{B}{n-1}\). Solving for \A\ and \B\, we find that \A = 1\ and \B = -1\, so: \(\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}\).

Substitute Back into the Series

Returning to the original summation: \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n(n-1)} = \sum_{n=2}^{\infty}((-1)^{n}\left(\frac{1}{n-1} - \frac{1}{n}\right))\).

Split and Rewrite the Series

Split the series into two separate sums: \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n-1} - \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n}\). Index-shift the first sum by taking \(k = n-1\) and adjusting the limits: \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\) and \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n}\).

Combine the Series

Rewrite each series with their common index form: \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} - \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n}\). Recognize that the first sum contains an additional term for \(n=1\).

Calculate the Final Sum

Extract the first term from the first sum and combine with the second: \(\frac{(-1)^{2}}{1} + \sum_{n=2}^{\infty} \left(\frac{(-1)^{n+1}}{n} - \frac{(-1)^{n}}{n}\right)\). Simplify to \(1\), since the remaining series cancels out.

Key concepts

Series Simplification

When dealing with series, especially alternating ones, simplifying the terms is crucial. Simplification involves breaking down the terms into more manageable expressions. In this exercise, the denominator of the series term \(\frac{(-1)^{n}}{n^{2}-n}\) is simplified first. Notice how \[n^{2}-n\] is factored into \[n(n-1)\]. This process makes it easier to handle each term separately, as it transforms a complex quadratic expression into simpler linear factors. Breaking down complex expressions helps us understand the behavior of each piece and makes further steps, such as partial fraction decomposition, more straightforward.

Partial Fraction Decomposition

Partial fraction decomposition is used to express a complicated fraction as the sum of simpler fractions. In our example, after simplifying the denominator to \[n(n-1)\], we decompose \[ \frac{1}{n(n-1)} \] into two fractions using the formula: \[ \frac{1}{n(n-1)} = \frac{A}{n} + \frac{B}{n-1} \]. To find \[A\] and \[B\], solve the identity for coefficients where \[1 = A(n-1) + Bn\]. This gives \[A=1\] and \[B=-1\], thus \[ \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} \]. Decomposition simplifies the overall series into a combination of simpler series, making it easier to manipulate and eventually sum.

Index Shift

The index shift in a series can simplify the expression by aligning the terms for easier combination. In our example, after partial fraction decomposition, the series is split into two parts: \[ \frac{(-1)^{n}}{n-1} - \frac{(-1)^{n}}{n} \]. For simplification, shift the index in the first series by letting \[ k = n-1 \]. This changes the summation limits and aligns the indices: \[ \frac{(-1)^{k+1}}{k} \]. An index shift helps by matching terms between split series, making it easier to identify and cancel out corresponding elements.

Series Combination

Combining series effectively can lead to significant simplification or cancellation of terms. After the index shift, the series in this exercise becomes: \[ \frac{(-1)^{k+1}}{k} - \frac{(-1)^{n}}{n} \]. Recognizing that the first sum includes the term \[n=1\], separate it out: \[ \frac{(-1)^{2}}{1} + \frac{(-1)^{n+1}}{n} - \frac{(-1)^{n}}{n}\]. The series from \[n=2\] onward cancels out, simplifying to yield the final sum of 1. Effective combination ensures that redundant terms are eliminated, and only essential terms are considered.