<s>[INST] What is wrong with this infinite sum We know that:
https://www.youtube.com/watch?v=w-I6XTVZXww
$$S=1+2+3+4+\cdots = -\frac{1}{12}$$
So multiplying each terms in the left hand side by $2$ gives:
$$2S =2+4+6+8+\cdots = -\frac{1}{6}$$
This is the sum of the even numbers
Furthermore, we can add it to itself but shifting the terms one place:
$$
\begin{align}
1+2+3+4+\cdots & \\ 1+2+3+\cdots & \\
=1+3+5+7+\cdots & =2S
\end{align}
$$
This is the sum of the odd numbers
If we were to now sum the odd numbers and the even numbers like below:
$$ 2+4+6+8+\cdots \\[6pt] 1+3+5+7+\cdots \\[6pt] \text{if we add the terms in a certain order we can get } 1+2+3+4+5+6+7+\cdots$$
This supposedly tells us that:
$$4S = S\\[6pt] 4 \left(\frac{-1}{12}\right)=\frac{-1}{12} \\[6pt] \frac{-1}{3} = \frac{-1}{12} $$
What is faulty with this proof.
 [/INST] Interpreted literally (i.e., using the usual sense of limits of infinite series), the first line,
$$1 + 2 + 3 + \cdots = -\frac{1}{12} ,$$
is simply false, as the series on the l.h.s. diverges.
What's true, for example, is that there's a natural way to extend the function $$Z(s) := \sum_{k = 1}^{\infty} k^{-s} ,$$ which is defined on $(1, \infty)$ (and in particular not at $s = -1$), to a function $\zeta$ defined for most complex numbers, including $s = -1$, and this function satisfies $\zeta(-1) = -\tfrac{1}{12}$. The partial sums of the series, evaluated at $s = -1$, are $1 + \cdots + n$, but this is not the same as saying $1 + 2 + 3 + \cdots = -\tfrac{1}{12}$.
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