What is wrong with this infinite sum

Question

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.

Answer 1

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 $1 + 2 + 3 + \cdots = -\tfrac{1}{12}$.

Answer 2

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.

Answer 3

As pointed out in section 3, page 1191 of this article, the rules for manipulating divergent series are more restrictive than those for convergent series. As pointed out in the article, to avoid problems you should work with the power series obtained by multiplying the nth term by $x^n$, instead.