<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] Lets try the same in a more general way: Instead of $-\frac{1}{12}$ I will use $C\in\mathbb{R}$.
So our infinite sum:
$$1+2+3+4+\cdots = C$$
Using the same technique you get these two equalities:
$$
2+4+6+8+\cdots = 2C
\\
1+3+5+7+\cdots = 2C
$$
And when you add them together:
\begin{align}
1+2+3+4+\cdots &= 4C
\\
C &= 4C
\end{align}
So for every nonzero real number $C$ you get $1 = 4$ which is obviously not true. So to answer your question: Why is the proof faulty for C = $-\frac{1}{12}$?? Because it's faulty for every nonzero real number. The mistake is, as it's been already said, in the first line. You can't prove something that is wrong. Learn more about this infinite series here.
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