Interesting sums

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


What is its value? Well, it turns out to be 12\frac{1}{2}...

So why is that true?

Well, let's call 11+11+11...=S1-1+1-1+1-1...=S.Then

1S=11+11+1...=S1-S=1-1+1-1+1...=S ! So 2.S=12.S=1 and S=12S=\frac{1}{2}

Now let's look at the sum

η(1)=12+34+56...=n=1n.(1)n1\eta(-1)=1-2+3-4+5-6...=\displaystyle\sum_{n=1}^{\infty} n.(-1)^{n-1} (The η(s)\eta(s) is the Dirichlet eta function )

Interestingly, the value of this sum is actually 14\frac{1}{4} !

Proof 1:We can rewrite the sum as


(11+11...)(1-1+1-1...) -

(11+11...)+(1-1+1-1...) +

(11+11...)...=(1-1+1-1...)- ...=


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

1+x+x2+x3...=11x1+x+x^2+x^3...=\frac{1}{1-x} for x less than 1.

Now differentiating both sides we get:

ddx(1+x+x2+x3...)=ddx11x\frac{\mathrm{d} }{\mathrm{d} x}(1+x+x^2+x^3...)=\frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{1-x}


Now substituting x=1x=-1 we get that


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

1+2+3+4...=n=1n1+2+3+4...=\displaystyle\sum_{n=1}^{\infty} n

Well, the sum is obviously divergent, but it's value can also be calculated as 112-\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:

ζ(s)=n=11ns=n=1ns\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 ζ(1)\zeta(-1).

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

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

2s.ζ(s)=2s.n=1ns=2s+4s+6s...=Ψ2^{-s}.\zeta (s) = 2^{-s}.\displaystyle\sum_{n=1}^{\infty} n^{-s}=2^{-s}+4^{-s}+6^{-s}...=\Psi

Now Υ2.Ψ=(12.2s)ζ(s)=\Upsilon-2.\Psi=(1-2.2^{-s})\zeta(s)=




That reminds us of the good ol'


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

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

So (3).ζ(1)=14(-3).\zeta (-1) = \frac{1}{4} and ζ(1)=112\boxed{\zeta (-1) = -\frac{1}{12}}

Note:The sum η(1)\eta(-1) is divergent, but Euler summable! The method we used to calculate ζ(1)\zeta (-1) is called analytic continuation which extends the domain of a given function. We sets=1s=-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
7 years, 1 month 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 this a fallacy?

Eddie The Head - 7 years, 1 month ago

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This is the video

It is not actually a fallacy! The sum 12+34+5...1-2+3-4+5-... , which we used to evaluate ζ(1)\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 14\frac{1}{4} , no rigorous explanation arrived until the mid 18-th century - Dirichlet eta function and Riemann zeta function

Bogdan Simeonov - 7 years, 1 month ago

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The grandi sum can also be evaluated to be equal to 0 and 1....

Eddie The Head - 7 years, 1 month ago

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@Eddie The Head Yes, but it should be 12\frac{1}{2}. See here

Bogdan Simeonov - 7 years, 1 month ago

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@Bogdan Simeonov But then why do we say that sum of integers is also an integer??

Satvik Golechha - 6 years, 11 months ago

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

G(t)=11+11+11+tG(t) = 1-1 + 1-1 + 1-1\ldots +t where tt is the last term of the series and t{1,1}t \in \{-1,1\}

We know that:

G(1)=1G(1) =1


G(1)+G(1)=1equation{1}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 SS to be a finite series in the proof given above:

Let SS have 2n+12n+1 terms, G(1)G(1) also has 2n+12n+1 terms while G(1)G(-1) has 2n2n terms.

S=11+11+11+1S = 1-1 + 1-1 + 1-1 \ldots +1\quad ( A series of (n+1)(n+1) 1s1's and nn (1)s(-1)'s. )

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

S=1(11+111)S = 1 - (1-1 +1-1 \ldots - 1)

S=1(G(1))S = 1-( G(-1)) \quad ( G(1)G(-1) is a series of nn 1s1's and nn (1)s(-1)'s. )

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

From equation{2} and equation{3},

G(1)+G(1)=1G(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 limn\displaystyle \lim_{n \to \infty} .

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

Thus, putting this in equation{3},


2S=1\Rightarrow 2S=1

S=12\Rightarrow \boxed{S=\frac{1}{2}}

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

Harsh Khatri - 5 years ago

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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 - 5 years 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 - 7 years, 1 month ago

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It is too complicated for Cosines Group (levels 2-3), and I have removed the tag.

Calvin Lin Staff - 7 years, 1 month 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 Staff - 7 years, 1 month ago

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That's strange... You could also evaluate V as zero, because it is equal to U-U...But it is also equal to ζ(0)\zeta(0), which I googled and found out is equal to 12-\frac{1}{2}

Bogdan Simeonov - 7 years, 1 month ago

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I don't follow proof 1.

Bob Krueger - 7 years, 1 month ago

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If you sum them vertically, it all makes sense.

Bogdan Simeonov - 7 years, 1 month ago

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