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Interesting sums

Firstly, let's look at the infinite sum (it's called Grandi's Series)

\(1-1+1-1+1-1...\)

What is its value? Well, it turns out to be \(\frac{1}{2}\)...

So why is that true?

Well, let's call \(1-1+1-1+1-1...=S\).Then

\(1-S=1-1+1-1+1...=S\) ! So \(2.S=1\) and \(S=\frac{1}{2}\)

Now let's look at the sum

\(\eta(-1)=1-2+3-4+5-6...=\displaystyle\sum_{n=1}^{\infty} n.(-1)^{n-1}\) (The \(\eta(s)\) is the Dirichlet eta function )

Interestingly, the value of this sum is actually \(\frac{1}{4}\) !

Proof 1:We can rewrite the sum as

\(1-2+3-4+5-6...=\)

\((1-1+1-1...) - \)

\((1-1+1-1...) + \)

\((1-1+1-1...)- ...= \)

\(S-S+S-S...=S^2=\frac{1}{4}\)

Proof 2:Let's look at the infinite geometric series:

\(1+x+x^2+x^3...=\frac{1}{1-x}\) for x less than 1.

Now differentiating both sides we get:

\(\frac{\mathrm{d} }{\mathrm{d} x}(1+x+x^2+x^3...)=\frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{1-x}\)

\(1+2x+3x^2+4x^3...=\frac{1}{(1-x)^2}\)

Now substituting \(x=-1\) we get that

\(1-2+3-4+5-6...=\frac{1}{4}\)

Now let's set our task to find the value of

\(1+2+3+4...=\displaystyle\sum_{n=1}^{\infty} n \)

Well, the sum is obviously divergent, but it's value can also be calculated as \(-\frac{1}{12}\) !!! Isn't it amazing? Well, maybe you think that I am some kind of lunatic, but nevertheless, I will show you the proof.

First, let me briefly introduct you to the Riemann zeta function.It is defined like this:

\(\zeta (s)=\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s} =\displaystyle\sum_{n=1}^{\infty} n^{-s}\)

So what we are seeking for is \(\zeta(-1)\).

The following observations were made by Leonhard Euler ( he didn't use the same notation):

\(\zeta (s) =\displaystyle\sum_{n=1}^{\infty} n^{-s}=1^{-s}+2^{-s}+3^{-s}+4^{-s}+5^{s}+6^{-s}...=\Upsilon\)

\(2^{-s}.\zeta (s) = 2^{-s}.\displaystyle\sum_{n=1}^{\infty} n^{-s}=2^{-s}+4^{-s}+6^{-s}...=\Psi\)

Now \(\Upsilon-2.\Psi=(1-2.2^{-s})\zeta(s)=\)

\(1^{-s}+2^{-s}+3^{-s}+4^{-s}+5^{s}+6^{-s}...\)

\(-2.2^{-s}-2.4^{-s}-2.6^{-s}...=\)

\(1^{-s}-2^{-s}+3^{-s}-4^{-s}+5^{-s}-6^{-s}...=\displaystyle\sum_{n=1}^{\infty}n^{-s}.(-1)^{n-1}=\eta(s)\)

That reminds us of the good ol'

\(\eta(-1)=1-2+3-4+5-6...=\frac{1}{4}\).

And indeed if we set \(s=-1\) we would get that

\((-3).\zeta(-1) =\eta(-1) \) !

So \((-3).\zeta (-1) = \frac{1}{4}\) and \(\boxed{\zeta (-1) = -\frac{1}{12}}\)

Note:The sum \(\eta(-1)\) is divergent, but Euler summable! The method we used to calculate \(\zeta (-1)\) is called analytic continuation which extends the domain of a given function. We set\(s=-1\), where the sum is divergent, but we used a formula for the values in the domain.Also, the whole post was inspired by Numberphile.

Note by Bogdan Simeonov
3 years, 8 months ago

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Numberphile is one of my fav channels too....can you post a link to the video? The series is divergent but stil we are getting a finite sum...is this a fallacy? Eddie The Head · 3 years, 8 months ago

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@Eddie The Head This is the video

It is not actually a fallacy! The sum \(1-2+3-4+5-...\) , which we used to evaluate \(\zeta(-1)\) is divergent, but using the summation method, we get that it is equal to a quarter! Euler admitted that this is a paradoxical equation.After Euler himself found out that the sum is equal to \(\frac{1}{4}\) , no rigorous explanation arrived until the mid 18-th century - Dirichlet eta function and Riemann zeta function Bogdan Simeonov · 3 years, 8 months ago

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@Bogdan Simeonov The grandi sum can also be evaluated to be equal to 0 and 1.... Eddie The Head · 3 years, 8 months ago

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@Eddie The Head Let the Grandi series be defined such that:

\(G(t) = 1-1 + 1-1 + 1-1\ldots +t \) where \(t\) is the last term of the series and \(t \in \{-1,1\} \)

We know that:

\(G(1) =1\)

\(G(-1)=0\)

\(G(1) + G(-1) = 1 \quad \ldots \ldots equation\{1\} \)

The value of the sum depends on the last term but since it is an infinite series, we cannot determine what the last term is.

Now, let us assume \(S\) to be a finite series in the proof given above:

Let \(S\) have \(2n+1\) terms, \(G(1)\) also has \(2n+1\) terms while \(G(-1)\) has \(2n\) terms.

\(S = 1-1 + 1-1 + 1-1 \ldots +1\quad \) ( A series of \((n+1)\) \(1's \) and \(n\) \((-1)'s\). )

\(\Rightarrow S = G(1) \) \(\quad \ldots \ldots equation\{1\} \)

\(S = 1 - (1-1 +1-1 \ldots - 1) \)

\(S = 1-( G(-1)) \quad \) ( \(G(-1)\) is a series of \(n\) \(1's\) and \(n\) \((-1)'s\). )

\(S + G(-1)=1 \quad \ldots \ldots equation\{3\} \)

From equation{2} and equation{3},

\(G(1) + G(-1)=1\) which is indeed true.

So, it is not a fallacy.

But a contradiction occurs when this is evaluated when applying the limit \(\displaystyle \lim_{n \to \infty} \).

This is where one commits the grave mistake of assuming S= G(-1).

Thus, putting this in equation{3},

\(S+S=1\)

\(\Rightarrow 2S=1 \)

\(\Rightarrow \boxed{S=\frac{1}{2}} \)

One feels it valid to substitute \( S = G(-1) \) because one cannot make out the difference when \(n \rightarrow \infty \) and so we assume that the sum does converge to a single value, indirectly saying \(G(1) = G(-1) \), which is not true.

Thus, it is not a fallacy just an invalid assumption.

The same can also be proved if we begin with

\( S = G(-1) \). :) Harsh Khatri · 1 year, 7 months ago

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@Eddie The Head Yes, but it should be \(\frac{1}{2}\). See here Bogdan Simeonov · 3 years, 8 months ago

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@Bogdan Simeonov But then why do we say that sum of integers is also an integer?? Satvik Golechha · 3 years, 6 months ago

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@Eddie The Head When I posted the problem of 1-1+1-1+1-1 ............... equal to 1/2 , my problem was deleted . I don't understand the reason if the value of 1/2 is recognized everywhere ? Raven Herd · 1 year, 7 months ago

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My roommate showed me a couple simpler proofs for 1 - 2 + 3 - 4... and 1 + 2 + 3 + 4. (Though, your proof for 1 - 2 + 3 - 4... was interesting. Your proof for 1 + 2 + 3 + 4 is above my level since it's been a while since I've properly studied summations.) I also don't know if the following proofs break the rules that define these infinite series in the first place, you'll have to call me out if that's the case. As you said:

S = 1 - 1 + 1 - 1 ...

Let's call the next series T

T = 1 - 2 + 3 - 4 ...

What's T + T? Let's add it vertically, with a shift:

T     = 1 - 2 + 3 - 4 ...
T     =     1 - 2 + 3 ...
T + T = 1 - 1 + 1 - 1 ... = S

So 2T = S, thus T=1/4.

Let's call the last sum, 1 + 2 + 3 + 4, U. The following is a strange trick, and I have no idea how somebody thought of it. What's U - 4U?

U      =  1  +  2  +  3  +  4 ...
4U     =        4        +  8 ...
U - 4U =  1  -  2   +  3  - 4 .... = T = 1/4
-3U    = 1/4
U      = -1/12

If this reasoning is wrong I'd love to know why. My friend and I were debating about it yesterday. Here's an example of what must be breaking some sort of rule:

V     = 1 + 1 + 1 + 1 + 1 + 1...
1 + V = 1 + (1 + 1 + 1 + 1...)
V     = V + 1
1     = 0

Where to draw the line on this? Dan Krol · 3 years, 8 months ago

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@Dan Krol That's strange... You could also evaluate V as zero, because it is equal to U-U...But it is also equal to \(\zeta(0)\), which I googled and found out is equal to \(-\frac{1}{2}\) Bogdan Simeonov · 3 years, 8 months ago

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Please inform me if this post is too complicated for Cosines Group or too easy for Torque group.Also, if you have any questions, ask me in the comments. Bogdan Simeonov · 3 years, 8 months ago

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@Bogdan Simeonov It is too complicated for Cosines Group (levels 2-3), and I have removed the tag. Calvin Lin Staff · 3 years, 8 months ago

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I don't follow proof 1. Bob Krueger · 3 years, 8 months ago

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@Bob Krueger \(1-2+3-4+5-6+7-8...=\)

\(1-1+1-1+1-1+1-1...\)

\(-1+1-1+1-1+1-1+1...\)

>>\(1-1+1-1+1-1+1-1...\)

>>>\(-1+1-1+1-1+1-1+1...\)

>>>>>\(\ldots\)

If you sum them vertically, it all makes sense. Bogdan Simeonov · 3 years, 8 months ago

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