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Chapter 5: Problem 43

Find the partial fraction decomposition for \(\frac{1}{x(x+1)}\) and use the result to find the following sum: $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{99 \cdot 100}$$

Short Answer

Expert verified
The sum of the series is \(\frac{99}{100}\).

Step by step solution

01

Perform the Partial Fraction Decomposition

Decompose the fraction \(\frac{1}{x(x+1)}\) into simpler fractions. Here we have two distinct linear factors in the denominator, so we can write as \(A/x + B/(x+1)\). Setting this sum equal to the original fraction, and clearing the denominator, we can equate the numerators to find \(A(x+1)+Bx=1\). Setting \(x=-1\) solves for A; setting \(x=0\) solves for B.
02

Solve for A and B

Now we solve the equations to obtain the values of A and B. When \(x=-1\), we get \(A(-1+1)=-A=1\) hence \(A=-1\). When \(x=0\), the equation becomes \(B*0=1\), hence \(B=1\). The fraction decomposes to \(\frac{-1}{x}+\frac{1}{x+1}\).
03

Apply Result to the Series Problem

We apply this decomposition to the given series: \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{99 \cdot 100}\) each term represents \(\frac{-1}{n}+\frac{1}{n+1}\), and hence modeled by our decomposition formula.
04

Compute the Series Sum

Observing that adjacent terms cancel each other (characteristic of a telescoping series) except the first term of the first fraction and the second term of the last fraction, we can sum up the series as: \(1 - \frac{1}{100}\)

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