\(\color{red}{\text{The following is a "proof" being circulated over the internet that claims the sum}\\ \text{of all the natural numbers is} \frac{1}{12} \text{.} \\ \\ \text{Can you figure out what is WRONG in this "proof"?}}\)
We'll consider three infinite series:
\[S_1 = 1  1 + 1  1 + 1  1 + \cdots\] \[S_2 = 1  2 + 3  4 + 5  6 + \cdots \] \[S_3 = 1 + 2 + 3 + 4 + 5 + 6 + \cdots \]
Note that when \(S_1\) contains an even number of terms, the sum is 0. When it contains an odd number of terms, the sum is 1. Since, it is either 0 or 1 with equal probability, \[S_1 = \frac{1}{2}(0+1) = \frac{1}{2}\]
Now, consider \(S_2\). We're going to add \(S_2\) to itself. When we write it, we'll do a bit of offset:
1 2 3 4 

\[\therefore 2S_2 = S_1 \\ \implies S_2 = \frac{1}{2} S_1 = \frac{1}{4}\]
Now, let's look at what happens if we take the \(S_3\), and subtract \(S_2\) from it:
1 2 3 4 

\[\therefore S_3  S_2 = 4 S_3 \\ \implies 3S_3 = S_2 \\ \implies S_3 = \frac{1}{3}S_2 \\ \implies S_3 = \frac{1}{12} \\ \boxed{\implies 1+2+3+\cdots = \frac{1}{12}}\]
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Top NewestCan you figure out what is WRONG in this "proof"?
I am searching for what is RIGHT in the "proof"...
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Oh god, why?
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How can one use probability for computing exact sums?
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If you have 2 hands you can use them . If you have 0 hands then you can not use them.So next time use your 1/2(0+2) or 1 hand................I am really searching for what is right in your proof.
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Do not search for right things. This proof is not mine, nor do I like it
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Yeah:
All the mathematical operations used here are improperly applied. The commutative property of addition only applies for well defined expressions, meaning, NOT those made up of an infinite amount of terms. A sequence cannot be manipulated in ways that Numberphile falsefully claims. For one, if you've seen the video, they differentiate power series and say that it applies for all \(r<1\), while we know that it applies only for \(r<1\) (and conveniently they let \(r=1\).
All the sequences above stated are not welldefined. For example:
\(11+11+...\) could be looked upon as \(1+(1+1)+(1+1)+...=1+0+0+...=1\), or, \((11)+(11)+(11)+...=0+0+0...=0\). Now yeah it makes sense to "average," right? But we just saw that both manipulations should be equally valid, but they're not.
Oh, and that bit of OFFSET isn't justified. But it can be, I've seen some other posts. Here's a better one:
Now to be honest I don't know much about the pipe I'm smoking here. Here's some wikia:
ss
The second key insight is that the alternating series 1 − 2 + 3 − 4 + · · · is the formal power series expansion of the function 1/(1 + x)2 with 1 substituted for x. Accordingly, Ramanujan writes:
s
Dividing both sides by −3, one gets c = −1/12. Generally speaking, it is dangerous to manipulate infinite series as if they were finite sums, and it is especially dangerous for divergent series. If zeroes are inserted into arbitrary positions of a divergent series, it is possible to arrive at results that are not selfconsistent, let alone consistent with other methods. In particular, the step 4c = 0 + 4 + 0 + 8 + · · · is not justified by the additive identity law alone. For an extreme example, appending a single zero to the front of the series can lead to inconsistent results.
Here's the full article: WIKI
Now I don't know much about maths, but I do know String Theory. And there, this sum is defined for 24dimensional strings. I wonder what mathematicians have to say about THAT.
I say, that it may converge to 1/12, but not equal to, just as the limit of \(\frac{\sin{x}}{x}\) as \(x\rightarrow 0\) is defined, but not the plugin value.
I'm out.
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Physicists use this as a result, since the concept of infinite sum seems divergent, but is actually convergent.
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Kaboobly Doo string theory,
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The problem here is that the given series are not convergent. That means they do not give a defenite number as answer. For example consider \( S_{1} \) . It is neither 0 or 1 as claimed as one cannot decide whether it contains an even number or odd number of terms. Same is the case with other two serieses . Mathematicaly such series are known as divergent serieses and the usual mathematical operations +,  etc cannot be applied to get WONDERFUL results like this. Anyway thanks to #Agnishom Chattopadhyay 16, India for mentining this problem where a majority of guys are less aware.
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