**Problem**: (Problem-Solving Through Problems) Let $ p $ and $ q $ be real numbers with $ \frac{1}{p}-\frac{1}{q} = 1 $, $ 0 < p \le \frac{1}{2} $. Show that

$ p+\frac{1}{2}p^2+\frac{1}{3}p^3+\cdots = q-\frac{1}{2}q^2+\frac{1}{3}q^3-\cdots $.

Solution: We're going to use a little bit of *gasp* calculus for this problem, because it's pretty nice. All we need is the following two facts:

$ -\ln{(1-x)} = x+\frac{1}{2}x^2+\frac{1}{3}x^3+\cdots $

and

$ \ln{(1+x)} = x-\frac{1}{2}x^2+\frac{1}{3}x^3-\cdots $,

which are simply Taylor Series centered at zero (Maclaurin Series).

We are given $ \frac{1}{p}-\frac{1}{q} = 1 $, which we can rewrite as $ (1+q)(1-p) = 1 $, or even better,

$ \ln{(1+q)}+\ln{(1-p)} = 0 $

$ -\ln{(1-p)} = \ln{(1+q)} $.

But by our infinite series representations above, we have

$ p+\frac{1}{2}p^2+\frac{1}{3}p^3+\cdots = q-\frac{1}{2}q^2+\frac{1}{3}q^3-\cdots $,

as desired. QED.

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Comment: It turns out that with the infinite series representations the problem is really simple, so the only real step is finding those. This is easily done by experimenting with the natural log function.

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Practice Problem: Prove that $ e^{i\theta} = \cos{\theta}+i\sin{\theta} $ using infinite series.

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