First, I'd to like say that I'm a beginner so could you answer simply please.
I'll explain the problem:
I was looking at the page 1+2+3+4+⋯ and found that the sum of the natural numbers equals $\frac{-1}{12}$.
So I remembered the case of a monotonic sequence. I started thinking about $u_n = u_{n-1} + 1$ with intial term $u_1 = 1$; normally, that's a monotonic sequence but at $\infty$ rank?
I don't know if I forgot some parameters, so please enlighten me!
Thanks.
The simplistic argument goes like this: By the binomial theorem $$ (1+x)^{-2} = \sum \binom{-2}{k}x^k = \sum\frac{(-2)^{\underline{k}}}{k!}x^k = \sum\frac{(-1)^k2^{\overline{k}}}{k!}x^k = \sum (-1)^k k $$ Here we use the falling power and rising power notations; $a^{\overline{n}} \equiv a(a+1)\cdots(a+n-1)$.
The above expansion is valid and convergent for $|x < 1|$ but the usual analytic continuation trick is to take it as being formally valid for all $x$. This says that formally, when $x=1$, $$ 1 - 2 + 3 -4 + 5 \ldots = (1+1)^{-2} = \frac{1}{4} $$ Now add to that series (again, formallly) $4$ times the series $1+2+3+4+\ldots$ and cleverly place the $n$-th term of the added series under the $2n$-th term of the alternating sign series: $$ -\frac{1}{4} + 4\times(1+2+3+4\ldots) = 1 + (-2+4) + 3 + (-4+8) + 5 + (-6+12) + \ldots \\ = 1+2+3+4+\ldots $$ And now solve algebraically to get $$1+2+3+4+\ldots = -\frac{1}{12} $$
Sheer sophistry? Perhaps, but the same sort of manipulations yield spectacular and important results when working with the analytic continuation of the Zeta function, so one should not write off such techniques out of hand.
