Just to offer an alternative solution, let
Let <em>F(x)</em> denote the antiderivative of <em>f(x)</em>. Then by the fundamental theorem of calculus, we can write, for instance,
When <em>x</em> = 1, all terms in the sum corresponding to <em>n</em> ≥ 1 vanish, so that <em>f</em> (1) = 1.
Integrating the series and interchanging the sum and integral (terms and conditions apply - see Fubini's theorem) gives
The constant <em>C</em> here corresponds exactly to <em>f</em> (1).
In the integral, substitute
so that it transforms and reduces to
Then we have
and by shifting the index to make the sum start at <em>n</em> = 1,
or equivalently,
Recall the Maclaurin expansion for ln(<em>x</em>) centered at <em>x</em> = 1, valid for |<em>x</em> - 1| < 1 :
By comparing to this series, we observe that the series in <em>F(x)</em> converges to
so long as |<em>x</em> ² - 1| < 1. Then
Differentiate both sides to recover <em>f(x)</em> = 1/<em>x</em> .