Sure, but if you see 1−2+3−4+5...=1/4, you already have all the intuition you need. The rest is detail.
The example I used, 1+2+4+8+…=−1, is the same kind of thing, a power series applied outside its domain of convergence, which I used instead because, while it doesn’t lend itself to the derivation directly, it looks more like the sum we seek (in particular, all positive integers on the left and a negative number on the right), and I expected the formula for an infinite geometric series to be more familiar to most readers.
Here is another explanation, kind of:
Taylor expansion of 1/(1+x)^2 is 1 − 2x + 3x^2 − 4x^3 + 5x^4...
When x = 1, it means that 1 − 2 + 3 − 4 + 5… = 1⁄4
But 1 − 2 + 3 − 4 + 5… can be written as 1 + 2 + 3 + 4 + 5… − 2×2 − 2×4 − 2×6...
= 1 + 2 + 3 + 4 + 5… − 2×(2 + 4 + 6...)
= 1 + 2 + 3 + 4 + 5… − 2×2×(1 + 2 + 3...)
= (1 − 2×2) × (1 + 2 + 3...)
= −3 × (1 + 2 + 3...)
So if 1 − 2 + 3 − 4 + 5… = 1⁄4, we get:
1⁄4 = −3 × (1 + 2 + 3...)
-1/12 = 1 + 2 + 3...
(Found here.)
Sure, but if you see 1−2+3−4+5...=1/4, you already have all the intuition you need. The rest is detail.
The example I used, 1+2+4+8+…=−1, is the same kind of thing, a power series applied outside its domain of convergence, which I used instead because, while it doesn’t lend itself to the derivation directly, it looks more like the sum we seek (in particular, all positive integers on the left and a negative number on the right), and I expected the formula for an infinite geometric series to be more familiar to most readers.