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Chapter 5: Problem 52

Find the partial fraction decomposition for \(\frac{2}{x(x+2)}\) and use the result to find the following sum: $$\frac{2}{1 \cdot 3}+\frac{2}{3 \cdot 5}+\frac{2}{5 \cdot 7}+\dots+\frac{2}{99 \cdot 101}$$

Short Answer

Expert verified
The sum of the series is \(1 - \frac {1}{101} = \frac{100}{101}\)

Step by step solution

01

Find the partial fraction decomposition

To find the partial fraction decomposition, the first thing to do is to express the given function as the sum of two fractions. In this case, we can express \(\frac{2}{x(x+2)}\) as \(\frac{A}{x} + \frac{B}{x+2}\). Now clear the denominator by multiplying throughout to obtain: \(2 = A(x+2) + Bx\). From this equation, evaluate the coefficients A and B by making x = 0 and x = -2.
02

Evaluate the Coefficients A and B

If we plug x = 0 into the equation obtained above, we get \(2 = 2A\), so \(A = 1\).\n\nIf we plug x = -2 into the equation, we get \(2 = -2B\), so \(B = -1\).\n\nTherefore, the partial fraction decomposition for \(\frac{2}{x(x+2)}\) is \(1/x - 1/(x + 2)\).\n\nWe can use this to easily compute the series.
03

Compute the series

Now, if we look at the sequence \(\frac{2}{1 \cdot 3}+\frac{2}{3 \cdot 5}+\frac{2}{5 \cdot 7}+\dots+\frac{2}{99 \cdot 101}\), we can see that the sequence represents the sum of the series \(\frac{2}{x(x+2)}\) where x ranges from 1 to 99.\n\nFrom step 2, \(\frac{2}{x(x+2)}\) = \(1/x - 1/(x + 2)\).\n\nSo our sequence is equivalent to \((1 - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + ... + (\frac{1}{99} - \frac{1}{101})\)\n\nHere we can observe a telescoping series where every term cancels out the preceding term, except for the first term of the first expression and the second term of the last expression.\n\nTherefore, the sum of the given sequence = 1 - 1/101

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