Prove that 1 = 2:

First, we prove that the series 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... is absolutely convergent:

Let A = sigma(1/n), n runs the integers not containing "9453" as substring

then A = (1 + 1/2 + 1/3 + 1/4 + ... + 1/9451 + 1/9452 + 1/9454 + 1/9455 + 1/9456 + ... + 1/10000) + (1/10001 + 1/10002 + 1/10003 + ... + 1/19451 + 1/19452 + 1/19454 + 1/19455 + 1/19456 + ... + 1/29451 + 1/29452 + 1/29454 + 1/29455 + 1/29456 + ... + 1/94528 + 1/94529 + 1/94540 + 1/94541 + 1/94542 + ... + 1/100000000) + (1/100000001 + 1/100000002 + 1/100000003 + ... + 1/100009451 + 1/100009452 + 1/100009454 + 1/100009455 + 1/100009456 + ... + 1/1000000000000) + ...

and each "()" has 9999, 9999^2, 9999^3, ... terms, respectively.

thus we have A < (1 + 1 + 1 + ... + 1) + (1/10000 + 1/10000 + 1/10000 + ... + 1/10000) + (1/100000000 + 1/100000000 + 1/100000000 + ... + 1/100000000) + ... (by direct comparison test, 1 to 1/10000 are all <= 1, 1/10001 to 1/100000000 are all < 1/10000, 1/100000001 to 1/1000000000000 are all < 1/100000000, etc.)

= 1*9999 + (1/10000)*9999^2 + (1/100000000)*9999^3 + ...

= 9999 + (9999^2)/10000 + (9999^3)/100000000 + ...

= 9999 / (1-9999/10000)

= 99990000 (a finite number)

thus A < 99990000, and A must be convergent.

Let B = sigma(1/n), n runs the integers containing "9453" as substring

then B = 1/9453 + 1/19453 + 1/29453 + ... + 1/89453 + 1/94530 + 1/94531 + 1/94532 + ...

and the first term of B is less than the first term of A (1/9453 < 1), the second term of B is less than the second term of A (1/19453 < 1/2), the third term of B is less than the third term of A (1/29453 < 1/3), etc.

Thus, by direct comparison test, B must be also convergent.

Hence A + B also converges (since sum of convergent series must also converge), however, A + B is 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ..., thus, we proved that 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... converges, thus 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... is absolutely convergent.

Besides, by the Mercator series: ln(1+x) = x - (x^2)/2 + (x^3)/3 + (x^4)/4 + (x^5)/5 - (x^6)/6 + ...

when x = 1, we have

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

= (1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + ...) - (1/2 + 1/4 + 1/6 + 1/8 + 1/10 + 1/12 + ...) (by the above section, 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... is absolutely convergent, thus, rearrangements do not change the value of the sum)

= (1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + ...) + ((1/2 + 1/4 + 1/6 + 1/8 + 1/10 + 1/12 + ...) - 2*(1/2 + 1/4 + 1/6 + 1/8 + 1/10 + 1/12 + ...))

= ((1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + ...) + (1/2 + 1/4 + 1/6 + 1/8 + 1/10 + 1/12 + ...)) - 2*(1/2 + 1/4 + 1/6 + 1/8 + 1/10 + 1/12 + ...)

= (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ...) - (2/2 + 2/4 + 2/6 + 2/8 + 2/10 + 2/12 + ...)

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

= 0

We proved that ln(2) = 0, and since ln(1) = 0, hence ln(2) = ln(1), besides, since the function ln is injective function in the interval (0,infinity), thus we have 2 = 1, i.e. 1 = 2, and the proof is done.