Fascinating Method for Finding Pi

From Jonas Castillo Toloza of Colombia comes this interesting method for finding pi...he has sent me numerous mathematical curiosities in the past...I just haven't the time to look at all of them carefully...

Using "triangular" number denominators, he contends that

pi - 2 = 1/1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - ...

Notice the two positive terms followed by two negatives, etc., something rather unusual...and "triangular" numbers are those which are generated by n(n+1) /2...

His proof goes soemthing like this...let A equal the sum of the odd terms and B be the sum of the even terms, that is:

A = 1/1 - 1/6 + 1/15 - ...
B = 1/3 - 1/10 + 1/21 - ...

Now A = 2/(1 * 2) - 2/(3 * 4) + 2/(5 * 6) - 2/(7 * + ...

which is also equal to

A = (2/1 - 2/2) - (2/3 - 2/4) + (2/5 - 2/6) - (2/7 - 2/ + ...

Now if we unite the terms with even denominators, he gets

- 1/1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/7 - ...

and that is equal to (- log 2) according to the well-known expansion

log (1 + x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - ....

The terms with the odd denominators is

2/1 - 2/3 + 2/5 - 2/7 + 2/9 - 2/11 + ...

that is equal to pi/2, according to a well-known expansion for pi...

therefore A = pi/2 - log 2.

Okay, now with B...here

B = 2/(2 * 3) - 2/(4 * 5) + 2/(6 * 7) - 2/(8 * 9) + ...

that is equal to

B = (2/2 - 2/3) - (2/4 - 2/5) + (2/6 - 2/7) - (2/8 - 2/9) + ...

If we unite the terms with even denominators we get

1/1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - ...

that he is equal to log 2.

The terms with odd denominators are equal to

- 2/3 + 2/5 - 2/7 + 2/9 - 2/11 + 2/13 - 2/15 + ...

that he is equal to pi/2 - 2

So therefore B = pi/2 - 2 + log .

Now let's put it together:

If we unite the two parts A and Bm we get

A + B = pi/2 - log 2 + pi/2 - 2 + log 2 = pi - 2.

Pretty neat, huh?