Chapter 1: Problem 14
\(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{2}-n}\)
Short Answer
Expert verified
The sum of the series is 1.
Step by step solution
01
Identify the Series
The given series is \(\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n^{2}-n}\).
02
Simplify the Denominator
Rewrite the denominator \(n^{2} - n\) as \((n)(n-1)\). Thus, the term becomes \(\frac{(-1)^{n}}{n(n-1)}\).
03
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: \(\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}\).
04
Substitute Back into the Series
Returning to the original summation: \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n(n-1)} = \sum_{n=2}^{\infty}((-1)^{n}\left(\frac{1}{n-1} - \frac{1}{n}\right))\).
05
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 \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n}\).
06
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 \(n=1\).
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.