Proof that $0=1$?

Question

I recently saw the following "proof" online, and couldn't pinpoint where the mistake was made:

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From a well known property, $$1+2+3+\cdots = -\frac{1}{12}.$$

Multiplying both sides by $-1,$ we get $$-1-2-3-\cdots = \frac{1}{12}.$$

We can thus rearrange these equations as follows:

\begin{align*} 1+2+3+4+\cdots= \, -\frac{1}{12} \\ -1-2-3-\cdots= \; \: \, \, \frac{1}{12} \\ -1-2-3-\cdots= \; \: \, \, \frac{1}{12} \\ 1+2+\cdots = -\frac{1}{12} \end{align*} Adding, the RHS clearly sums to $0$, while the LHS yields $1$, seemingly yielding that $0=1$. Where did this proof go wrong?

Answer 1

The "well-known property" that you mention in the beginning does not hold. Therefore, neither does anything that you state after that.

Answer 2

The “well known” property is derived from Ramanujan summations, in which divergent sums are treated differently than normal, but then you treat the divergent sum in a typical manner. Plus being able to rearrange the terms and get the same sum is something reserved for absolutely convergent series, which the LHS is not.

Answer 3

Well, at least we have that $$\sum_{n \in \mathbb{N}} n=-\frac{1}{12} \implies 1=0$$ Is true.

Answer 4

I think a lot of people have been confused by the "well known property" that you are mentioning, so I am going to elaborate a little bit.

First of all, it is not true that $$ 1+2+3+\dots = -\frac{1}{12}. $$ And the above shouldn't be true. After all, it doesn't make any sense!

I will briefly explain why you see this identity in a lot of places. Basically, there is a function known as the zeta function which is of particular importance in number theory. For all numbers $s>1$, the zeta function is given by the formula $$ \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ If you were to pick $s\leq 1$, the the above sum would be divergent and $\zeta$ would not be well defined.

On the other hand, the function $\zeta$ can be extended to the real line in a meaningful way (this is known as analytic continuation and if taught in undergraduate complex analysis courses). So, we have function $\zeta$ which is defined for all numbers $s$ and such that $$ \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ whenever $s>1$.

Now, it turns out that $\zeta(-1) = -1/12$. Plugging $s=-1$ into $$ \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ one would obtain $$ 1+2+3+\dots $$ This is where your "well known identity" comes from.

Answer 5

I couldn't pinpoint where the mistake was made...

In order to understand the original fake proof, I suggest you start with this one, which is simpler:

From a well known property, $$1+2+3+\cdots = \infty.$$

Multiplying both sides by $-1,$ we get $$-1-2-3-\cdots = -\infty.$$

We can thus rearrange these equations as follows:

\begin{align*} 1+2+3+4+\cdots&= \infty\\ \\ -1-2-3-\cdots&= -\infty\\ \\ -1-2-3-\cdots&= -\infty\\ \\ 1+2+\cdots &= \infty \end{align*} Adding, the RHS clearly sums to $0$, while the LHS yields $1$, seemingly yielding that $0=1$.

Can you find the mistake for this one?

Answer 6

You start from a Ramanujan summation. In the next steps you use linearity (multiplication with $-1$ as well as termwise addition), stability (extracting finitely many summands is allowed), and regularity (for the convergent series of all $0$ terms, Ramanujan summation yields $0$). Can you justify that Ramanujan summation is linear, stable, and regular?

There are summation methods that are stronger than standard summation (=limit of partial sum), for example Cesàro summation. However, none of these assigns a finite value to $\sum n$, and your argument shows exactly the reason why a finite value is not possible for such a summation method.