Suppose you take 1/A, and subtract 1/B from it. Well, it gets smaller, right? But there's another way to make 1/A smaller: increase A. Specifically, replace A by A + x. If you choose x shrewdly, you can make 1/(A + x) = 1/A - 1/B. What x does the trick?
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Well, it's no deep mystery. Since 1/A - 1/B = (B - A)/AB, we want A + x = AB/(B - A). So x = (A^2)/(B - A). Problem solved! But the astonishing thing is, we can just keep going. Next, replace 1/B by 1/B - 1/C. That'll take us from 1/A - 1/B to 1/A - 1/B + 1/C.
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But remember that 1/B - 1/C = 1/(B + y), where y = (B^2)/(C - B), just as before. So 1/A - 1/B + 1/C = 1/A - 1/(B + y), which means that we can just replace B by B + y in 1/(A + x), to get an equivalent form for 1/A - 1/B + 1/C.
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And now it's clear there's no stopping us. We next replace 1/C by 1/C - 1/D = 1/(C + z), and repeat. And so on, and so forth, ad infinitum. We get a continued fraction from an alternating series of reciprocals. This is how Euler derived Brouncker's continued fraction for 4/pi.
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