Euler's proof of sum of natural numbers I've to recheck Euler's proof of the sum of the natural numbers, but I don't know exactly what it is? It has something to do with the $\zeta(s)$?
 A: Euler was assuming the following things:
$$0+1+x+x^2+x^3+x^4+...=\sum_{k=0}^{\infty} x^k=\frac{1}{1-x},|x|<1$$
If we take the $\frac{d}{dx}$ (derivative) of that sum we get:
$$0+1+2x+3x^2+4x^3+5x^4+...=\sum_{k=0}^{\infty} (k+1)x^k=\frac{1}{(1-x)^2},|x|<1$$
Second thing he assumed was:
$$\sum_{k=0}^{\infty} (k+1)x^k=1-2+3-4+5-6+...=\frac{1}{4}$$

Eulers proof of the sum of all natural numbers:
$$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...\Longleftrightarrow$$
$$2^{-s}\cdot \zeta(s)=2^{-s}\cdot \left(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...\right)\Longleftrightarrow$$ 
$$2^{-s}\cdot \zeta(s)=2^{-s}+4^{-s}+6^{-s}+8^{-s}+...\Longleftrightarrow$$ 
$$(1-2\cdot 2^{-s})\zeta(s)=(1-2\cdot 2^{-s})\left(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+...\right)\Longleftrightarrow$$
$$(1-2\cdot 2^{-s})\zeta(s)=1+2^{-s}+3^{-s}+...-2(2^{-s}+4^{-s}+6^{-s})\Longleftrightarrow$$
$$(1-2\cdot 2^{-s})\zeta(s)=1-2^{-s}+3^{-s}-4^{-s}+...$$
Now we choose $s=-1$:
$$(1-2\cdot 2^{--1})\zeta(-1)=1-2^{--1}+3^{--1}-4^{--1}+...\Longleftrightarrow$$
$$(1-2\cdot 2^1)\zeta(-1)=1-2^1+3^1-4^1+...\Longleftrightarrow$$
$$(1-2\cdot 2)\zeta(-1)=1-2+3-4+...\Longleftrightarrow$$
$$-3\cdot\left(\sum_{n=1}^{\infty} \frac{1}{n^{-1}}\right)=1-2+3-4+...\Longleftrightarrow$$
$$-3\cdot\left(\sum_{n=1}^{\infty} n\right)=1-2+3-4+...\Longleftrightarrow$$
$$-3\cdot\left(1+2+3+4+5+...\right)=1-2+3-4+...\Longleftrightarrow$$
$$-3\cdot\left(1+2+3+4+5+...\right)=\frac{1}{4}\Longleftrightarrow$$
$$1+2+3+4+5+...=\frac{\frac{1}{4}}{-3}\Longleftrightarrow$$
$$1+2+3+4+5+...=-\frac{1}{12}\Longleftrightarrow$$
$$\sum_{n=1}^{\infty} n=-\frac{1}{12}$$
