# The sum of all natural numbers = -1/12

Discussion in 'Physics & Math' started by Magical Realist, Feb 6, 2014.

1. ### RJBeeryNatural PhilosopherValued Senior Member

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4,222
Agreed, but I gave two methods of deducing that $\sum_{k=1}^{\infty} (1) = 0$ that did not use grouping. I was wondering if you knew for certain that the answer was 1/2 (or -1/2?)

3. ### rpennerFully WiredValued Senior Member

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4,833
I don't see how your methods are anything other than rearrangement. Since these series are divergent, they can be improperly arranged to resemble any infinite series.
$C = \sum_{k=1}^{\infty} (-1)^{k+1} \\ K = \sum_{k=1}^{\infty} 1$
Both series are divergent, but let's ignore that and see how we can attempt to assign sums to them.
Geometric Series
$\sum_{k=1}^{\infty} a_0 r^{k-1} = \frac{a_0}{1 - r} \quad \; \quad |r| < 1 \\ C \to \frac{1}{2}$
Equivalently:
$C = 1 - C \quad \rightarrow \quad 2 C = 1 \quad \rightarrow \quad C = \frac{1}{2}$
(C, 1) Cesàro summation
$C \to \lim_{n\to\infty} \frac{\sum_{j=1}^{n} \sum_{k=1}^{j} (-1)^{k+1}}{n} = \lim_{n\to\infty} \frac{\sum_{j=1}^{n} \frac{1 - (-1)^j}{2}}{n} = \lim_{n\to\infty} \frac{2 n + 1-(-1)^n }{4 n} = \frac{1}{2}$
Equivalently:
$C \to \lim_{n\to\infty} \sum_{j=0}^{n} \frac{n! \, (n+1-j)!}{(n+1)! \, (n-j)!} (-1)^j = \lim_{n\to\infty} \sum_{j=0}^{n} \frac{n+1-j}{n+1} (-1)^j = \lim_{n\to\infty} \frac{2n + 3+(-1)^n}{4n + 4} = \frac{1}{2}$
(C, 2) Cesàro summation
$C \to \lim_{n\to\infty} \sum_{j=0}^{n} \frac{n! \, (n+2-j)!}{(n+2)! \, (n-j)!} (-1)^j = \lim_{n\to\infty} \frac{2 n^2 + 8n + 7 + (-1)^n}{4n^2 + 12n + 8} = \frac{1}{2}$
Abel summation
$C \to \lim_{\epsilon\to 0^+} \sum_{k=1}^{\infty} (-1)^{k+1} e^{-(k-1) \epsilon} = \lim_{\epsilon\to 0^+} \frac{e^\epsilon}{1 + e^\epsilon} = \frac{1}{2}$
Lindelöf summation
$C \to \lim_{x \to 0} \sum_{k=1}^{\infty} (-1)^{k+1} k^{-k x} = \frac{1}{2}$
Euler summation
$C \to \sum_{k=0}^{\infty} ( 1 + x )^{-1-k} \sum_{j=0}^{k} \frac{ (-1)^j k! x^{j+1} }{ j! (k-j)! } = \sum_{k=0}^{\infty} \frac{x (1-x)^k}{(1 + x )^{1+k}} = \lim_{n\to\infty} \frac{1-\left( \frac{1-x}{1+x} \right)^{n+1}}{2} = \frac{1}{2}$
Zeta function regularization (Analytic continuation)
$C \to (1 - 2) \zeta(0) = \frac{1}{2} K \to \zeta(0) = -\frac{1}{2}$
Borel summation
$C \to \lim_{t \to \infty} e^{-t} \sum_{j=0}^{\infty} \frac{t^j}{j!} \sum_{k=1}{j} (-1)^{k+1} = \lim_{t \to \infty} \frac{ e^(2 t) - 1}{2 e^{2t} } = \frac{1}{2}$
Ramanujan summation
$K \to -\frac{1}{2}$

5. ### RJBeeryNatural PhilosopherValued Senior Member

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4,222
Hmm, that's certainly a lot of TEX, but restricting this to only the summation in question we're left with:

...which I find to be more like assertions than proofs.

I give three reasons for my claim that K = 0:

1) $S_1 = 1 + 2 + 3 + 4 + 5 + ... = -1/12$ and $S_1 - S_1 = 0$ but also

Code:
   1 + 2 + 3 + 4 + 5 + ...
-      1 + 2 + 3 + 4 + ...
---------------------
1 + 1 + 1 + 1 + 1 + ...

2) I agree with you on the "grouping" objection (please note, there was NO grouping above, only a shifting), because the graph in post #37 would change under grouping; basically grouping affects the X-axis values. Using the graphing analogy in #37, however, we can see that 1 + 1 + 1 + 1 + 1 + ... would correlate to plots points of (1, 1), (2, 2), (3, 3), (4, 4), (5, 5)...and that graph would clearly have a Y-axis intercept of zero.

$1 + 2^{2k} + 3^{2k} + \cdots = 0\ (\Re)$

If you set (k = 0) then we have a 3rd and more definitive proof that K = 0.

7. ### RJBeeryNatural PhilosopherValued Senior Member

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4,222
Extending the concept that "shifting is OK but grouping is not", we have
Code:
   1 + 2 + 3 + 4 + 5 + ...
-  0 + 1 + 2 + 3 + 4 + ...
---------------------
1 + 1 + 1 + 1 + 1 + ...

1 + 2 + 3 + 4 + 5 + ...
- -1 + 0 + 1 + 2 + 3 + ...
---------------------
2 + 2 + 2 + 2 + 2 + ...

1 + 2 + 3 + 4 + 5 + ...
- -2 +-1 + 0 + 1 + 2 + ...
---------------------
3 + 3 + 3 + 3 + 3 + ...


So $K_1 = \sum_{k=1}^{\infty} 1 = K_2 = \sum_{k=1}^{\infty} 2 = K_3 = \sum_{k=1}^{\infty} 3 {...} = 0$, which is consistent with both points #1 and #2 above.

8. ### rpennerFully WiredValued Senior Member

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4,833
The same page requires that 2k be positive.
The correct way to do a Ramanujan summation [old = C(0), new = C(1) ] is as follows:

$f(n) = 1 , \; f'(n) = 0, \; f^{(3)}(n) = 0, \; f^{(5)}(n) = 0, \cdots \begin{eqnarray} C(a) & = & \int_0^a f(t) \,dt & - & \frac{1}{2} f(0) & - & \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!}f^{(2k-1)}(0) \\ & = & a & - & \frac{1}{2} \times 1 & - & \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!}\times 0 \\ & = & a & - & \frac{1}{2} & - & 0 \\ & = & a - \frac{1}{2} \end{eqnarray} \sum_{k=1}^{\infty} f(k) = C(\infty) = \infty \sum_{k\geq 1}^{\Re} f(k) = C(0) = - \frac{1}{2} \sum_{k\geq 1}^{\Re'} f(k) = C(1) = \frac{1}{2}$ $f(n) = b^n , \; f'(n) = b^n \ln b, \; f^{(3)}(n) = b^n \left( \ln b \right)^3, \; f^{(5)}(n) = b^n \left( \ln b \right)^5, \cdots \begin{eqnarray} C(a) & = & \int_0^a f(t) \,dt & - & \frac{1}{2} f(0) & - & \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!}f^{(2k-1)}(0) \\ & = & \int_0^a b^t \,dt & - & \frac{1}{2} \times 1 & - & \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!}\times \left( \ln b \right)^{2k-1} \\ & = & \frac{b^a - 1}{\ln b} & - & \frac{1}{2} & - & \left( \frac{b + 1}{2 ( b - 1) } - \frac{1}{\ln b} \right) \\ & = & \frac{b^a \left( b - 1 \right) - b \ln b}{(\ln b)(b-1)} \end{eqnarray} \sum_{k=1}^{\infty} f(k) = C(\infty) = \infty \quad \quad \quad ; b > 1 \sum_{k\geq 1}^{\Re} f(k) = C(0) = \frac{b ( 1 - \ln b ) - 1}{(\ln b)(b-1)} \sum_{k\geq 1}^{\Re'} f(k) = C(1) = \frac{b \left( b - 1 \right) - b \ln b}{(\ln b)(b-1)}$

This method is self-consistent in that $\lim_{b \to 1} \frac{b ( 1 - \ln b ) - 1}{(\ln b)(b-1)} = -\frac{1}{2} , \quad \lim_{b \to 1} \frac{b \left( b - 1 \right) - b \ln b}{(\ln b)(b-1)} = \frac{1}{2}$

Last edited: Feb 19, 2014
9. ### RJBeeryNatural PhilosopherValued Senior Member

Messages:
4,222
I found it curious that you arrived at a different answer so I kept digging. It appears my methods were problematic on a level which I cannot appreciate, something to do with stable and linear methods. Anyway this page should erase all doubt (not that I should've doubted rpenner's math skills in the first place...) It also explains why my Y-axis intercept is not actually zero, so I feel better.

I concede.

10. ### eramSciengineerValued Senior Member

Messages:
1,877
So the solution to this "paradox" is that the divergent series are improperly arranged?

11. ### rpennerFully WiredValued Senior Member

Messages:
4,833
No -- the solution is that convergent series, like $\sum_{k \geq 1} \frac{1}{2^k}$ cannot have their value changed by grouping or rearrangement that moves no element more than a finite amount, while divergent series, like $\sum_{k \geq 1} \frac{1}{k}$ or $\sum_{k \geq 1} (-1)^{k+1}$ do not have a normal value, but can have a regularized value via any number of methods but this value depends critically on the arrangement of terms.

Trivially, each of the following purported algebraic manipulations has a problem:
$\sum_{k \geq 1} (-1)^{k+1} = \sum_{k = 1}^{\infty} (-1)^{k+1} =^? \sum_{k = 1}^{\infty} (-1)^{2k} + (-1)^{2k+1} = \sum_{k = 1}^{\infty} ( 1 - 1 ) = 0 \sum_{k \geq 1} (-1)^{k+1} = \sum_{k = 1}^{\infty} (-1)^{k+1} = 1 + \sum_{k = 2}^{\infty} (-1)^{k+1} =^? 1 + \sum_{k = 1}^{\infty} (-1)^{2k+1} + (-1)^{2k+2} = 1 + \sum_{k = 1}^{\infty} ( -1 + 1 ) = 1 + 0 = 1 \sum_{k \geq 1} (-1)^{k+1} = 1 + \sum_{k \geq 2} (-1)^{k+1} = 1 + \sum_{k \geq 1} (-1) (-1)^{k+1} =^? 1 + (-1) \sum_{k \geq 1} (-1)^{k+1} = \frac{1}{2}$
Only by adopting a particular regularization scheme can one find an answer without invalid algebra, but such an answer is not the same as the limit of the sum as the number of terms grows without bound -- that limit does not exist in this case, therefore the series is divergent.