 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2004/d/d1/solution

For

x > 0

the geometric ratio

e^{- 2 x}

has modulus less than one, and

\sum_{m \geq 0} (1 - e^{- x}) e^{- (2 m + 1) x} = \frac{( 1 - e ^{- x} ) e ^{- x}}{1 - e ^{- 2 x}} = \frac{1}{e ^{x} + 1} .

Every summand is nonnegative. The monotone convergence theorem therefore permits integration of the sum:

\int_{0}^{\infty} \frac{d x}{e ^{x} + 1} = \sum_{m \geq 0} (\frac{1}{2 m + 1} - \frac{1}{2 m + 2}) .

The integral is finite because its integrand is bounded above by

e^{- x}

, whose integral is one. The displayed paired sums are the even partial sums of the alternating harmonic series; its odd partial sums differ by a term tending to zero. Thus the required equality holds with the usual series ordering. One can additionally evaluate it by

t = e^{- x}

\int_{0}^{\infty} \frac{d x}{e ^{x} + 1} = \int_{0}^{1} \frac{d t}{1 + t} = lo g 2 = n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n} .

Grouping into nonnegative pairs supplies the justified interchange; absolute convergence of the ungrouped integrated alternating series is not needed and would be false.

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