The sum of all the odd numbers to infinity

9,986

Solution 1

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 $\infty$.

When dealing with divergent series, as with quantum mechanics, things are not always the way one would expect.

Solution 2

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.

Solution 3

S1 is divergent series. It doesn't have a finite sum and this possibility of rearranging terms in order to get different sums is a consequence of that. An issue should be immediately apparent once you see a monotonically increasing series of non-negative terms converging to a negative number.

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Updated on August 10, 2022

Comments

  • It'sRainingMen
    It'sRainingMen about 1 year

    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 ∞