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Leonhard Euler's sums


Euler began his proof by introducing the function

To Euler, f (x) was just an infinite polynomial with f (0) = 1. Thus, it can be factored provided the roots of the equation f (x) = 0 are determined. Keeping in mind that x is not equal to zero:

, by the Taylor Expansion of sin x

Then, as long as x is not 0, solving f (x) amounts to solving 0, which by using a simple cross-multiplication, reduces to solving sin (x) = 0. The sine function equals 0 precisely for x = 0, x = + /- Pi, x = +/- 2Pi..... But we have to eliminate x = 0 from consideration as a solution to f (x) = 0, since we have already noted that f (0) = 1. Using these considerations, Euler factored f (x) as:


which amounts to


This is a pretty cool result because it changes an infinite sum with an infinite product. Which means that the infinite series by which f (x) was originally defined has been equated to the infinite product on the right. Euler then went on by multiplying out the infinite product on the right side of the preceding equation and then collecting all terms having the same power of x. In doing so, the first term to appear would be the product of all of the 1s (which turns out to be 1). To end up with an term, we would have to multiply the 1s from all but one of the factors by an tern from that remaining factor. Then Euler's infinite multiplication problem would give us this equation:



After Euler had multiplied out the infinite product to get two infinite sums equal to each other, he went on to equate the like powers of x. Next comes the term in each series, and so the coefficients must be equal, which is:

Then multiplying both sides by -1, and making 3! = 6 on the left side, and factoring the common , Euler arrived at

and then a final cross multiplication yielded the astonishing fact that


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