# The sum of all the odd numbers to infinity [duplicate]

We have this sequence:

S1: 1+2+3+4+5+6.. (to infinity)

It has been demonstrated, that S1 = -1/12.

Now, what happens if i multiply by a factor of 2?

S2: 2+4+6+8+10+12.... (to infinity).

I have 2S1, which is equal to -1/6

On this, we can create a equation for the odd numbers:

S3: 1+3+5+5+7+9+11... (to infinity)

We know that for every term in S2, every term in S3 is just (n-1)

Or, The sum of the even numbers, Minus , the sum of infinitely many (-1)s

So S3 = -1/6 - ∞

However, we also know that the odd numbers + the even numbers = The natural numbers.

So let's try it.

-1/6 - (-1/6 -∞)

We have -1/6 + 1/6 + ∞

Which is just ∞

So, there we have it. a paradox. S1 cannot be both -1/12 or ∞

## marked as duplicate by AlexR, Hakim, amWhy, Hayden, Jean-Claude ArbautOct 24 '14 at 13:13

• This $1+2+3+4....=\frac {-1} {12}$ thing is an old gig, and is not true in for the usual definition of a summation of a sequence. – Alan Oct 24 '14 at 11:20
• $\sum_{n=1}^\infty a_n$ doesn't converge to $-1/12$. It diverges to infinity. There are various, very complicated, mathematical reasons to say things like "if $\sum_{n=1}^\infty a_n$ were to converge, then it must be to $-1/12$". But, writing $\sum_{n=1}^\infty a_n = -1/12$ is somewhat misleading, because you aren't using the definition of convergent series. – James Oct 24 '14 at 11:21
• So you showed by manipulating infinite divergent series like they were convergent that you cannot manipulate infinite divergent series like they were convergent. – UserX Oct 24 '14 at 11:23
• @UserX, this is very nice comment – Galc127 Oct 24 '14 at 11:29
• @AlexR: I don't think that is the confusion; it is the manipulations after that that are causing the confusion. The odd part is that this was closed as a duplicate about 50 minutes before I posted my answer. – robjohn Oct 24 '14 at 14:40

Since $$\zeta(z)=\sum_{k=1}^\infty\frac1{k^{\large z}}$$ and $$2^{-{\large z}}\zeta(z)=\sum_{k=1}^\infty\frac1{(2k)^{\large z}}$$ we have $$\left(1-2^{-{\large z}}\right)\zeta(z)=\sum_{k=1}^\infty\frac1{(2k-1)^{\large z}}$$ Thus, the same reasoning that says $$\sum_{k=1}^\infty k=\zeta(-1)=-\frac1{12}$$ would say that $$\sum_{k=1}^\infty2k=2\zeta(-1)=-\frac16$$ and $$\sum_{k=1}^\infty(2k-1)=(1-2)\zeta(-1)=\frac1{12}$$ Taking the difference of these last two, would lead one to conclude that $$\sum_{k=1}^\infty1=-\frac14$$ not $\infty$.
If you start by accepting that in some sense $1+2+3+\ldots=-\frac1{12}$ instead of diverging, you should also not assume a priori that $(-1)+(-1)+(-1)+\ldots =-\infty$. Instead, for a consistant summation method that assigns values to $S_1=1+2+3+\ldots$ and to $S_3=1+3+5+\ldots$ and to $S_2=2+4+6+\ldots$ and that allows the manipulations you used (multiply by a constant factor, add summandwise, partition), we must have $1+1+1+\ldots = 2S_2-S_3$ and $S_2=2S_1$ and $S_1=S_2+S_3$, hence $S_3=-S_1$ and $1+1+1+\ldots = 5S_1$ ($\ne\infty$ if $S_1\ne\infty$). You can't use the standard summation (limit of partial sums) in one part of your argument and another summation in other parts.