1+2+3+4+5+6+7+... = 3/16

Let S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + ...

If we isolate the odd and even numbers, we obtain

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

Let's start with the odd terms. Collect the terms in groups of 5 consecutive odd numbers:

(1+3+5+7+9) + (11+13+15+17+19) + (21+23+25+27+29) + ...

= 25 + 75 + 125 + 175 + 225 + 275 + ...

= 5*5 + (50 + 5*5) + (100 + 5*5) + (150 + 5*5) + ...

We can separate all the multiples of 50:

= (50 + 100 + 150 + 200 + 250 + ...) + (5*5 + 5*5 + 5*5 + 5*5 + 5*5 + ...)

= 50 (1+2+3+4+5+...) + (5*5 + 5*5 + 5*5 + 5*5 + 5*5 + ...)

= 50(S) + (5+5+5+5+5) + (5+5+5+5+5) + (5+5+5+5+5) + ...

Note: we obtain a single 5 for every odd term in the sum.

Now we still have the even terms remaining. Since we have an equal number of even terms and 5s from previously, we can add them together pair-wise:

(2 + 4 + 6 + 8 + 10 + ...) + (5 + 5 + 5 + 5 + 5 + ...)

= (2 + 5) + (4 + 5) + (6 + 5) + (8 + 5) + (10 + 5) + ...

= 7 + 9 + 11 + 13 + 15 + ...

Note: this sum is simply the sum of all odd numbers, minus the first 3 terms.

The sum of all odd numbers can be written as the sum of all numbers minus the sum of all even numbers:

1 + 3 + 5 + 7 + ... = (1 + 2 + 3 + 4 + 5 + 6 + ...) - 2 - 4 - 6 - 8 - .... = S - 2(1 + 2 + 3 + 4 + ...) = S - 2S

Combined with the result above, we get 7 + 9 + 11 + 13 + 15 + ... = S - 2S - (1 + 3 + 5) = S - 2S - 9

Putting it together we have:

S = 50S + (S - 2S) - 9

S = 49 S - 9

-48 S = -9

S = 9/48 = 3/16

:)