-1/12 = oo, for small values of oo

Having too much time on my hands …

A few months ago, there was a minor Internet Brouhaha over a video (hereafter referred to as the first video) which proports to show that

1 + 2 + 3 + 4 + 5 + … = − 1 12   .

(1)

This was picked up by Phil Plait, the Bad Astronomer, and was followed up by a slightly more rigorous proof, https://www.youtube.com/watch?v=E-d9mgo8FGk ("the second video"), some backtracking/correcting, a Wikipedia entry, and a lot of web pages.

So, is equation (1) really true?

Not really, but kinda-sorta.

It is true that the sum

∞ ∑ n=1  n

(2)

diverges, but we can approach the result (1) by several different lines of attack.

The standard method, which is referred to in the second video quoted above, starts with the Riemann Zeta Function:

ζ(s) = ∞ ∑ n=1  1 ns   ,   ℜs > 1   .

(3)

This series only converges, as noted, when the real part of s is greater than one, but it can be analytically continued (a process that I understand in principle but don't know how to put into practice here) to give you results for other values of s. In particular, it is possible to show (see equation 25.6.3) that

ζ(−1) = − 1 12   .

(4)

Now if you just set s = -1 in (3) it looks like you get equation (1). This isn't exactly true, but we can say that

∞ ∑ n=1  n → ζ(−1) = − 1 12   ,

(5)

where, in this case, → means by a procedure whose description will not fit in the margin of this web page.

That's one way to do it. I'm going to go at it another way. It's akin to the procedure that is described in the first video, but, I hope, more rigourous, though I have probably missed a few steps. It may be equivalent to Ramanujan summation, but don't quote me on that.

Let's start with the series

S1(x) = ∞ ∑ n=0  xn   ,   |x| < 1   .

(6)

This is well defined so long as x is between -1 and 1, and we all know that we can also write

S1(x) = 1 1−x   .

(7)

This equation gives a finite value for all x, except the pole at x = 1, and is equal to (6) for |x| < 1. So we can justifiably say that (7) is the analytic continuation of (6). Also note that if we put x = −1 into (6) and evaluate it by using (7) we get the result

S1(−1) → 1 − 1 + 1 − 1 + 1 − 1 + …→ 1 2   ,

(8)

where we use the arrow rather than an equals sign to show that these results are obtained only in the limit where x → −1 from above. The result (8) should be familiar from the first video.

Now look at the derivative of (6) and (7). We'll call this S₂:

S2(x) = S1′(x)   .

(9)

Differentiating (7) we get

S2(x) = 1 (1−x)2   ,

(10)

which is defined over the entire complex plane except at x = 1. We can also differentiate (6) and get

S2(x) = ∞ ∑ n=1  n  xn−1   ,   |x| < 1.

(11)

Again, we can analytically continue (11) into (10) to define the function over the entire complex plane, except for the pole at x = 1. And again, from (10) we have

S2(−1) = 1 4   ,

(12)

while from (11) we get

S2(−1) → 1 − 2 + 3 − 4 + 5 − 6 + …  .

(13)

This, too, was shown in the first video.

Now let's take a look at the difference between S₂(x) and S₂(−x). Define

F2(x) = S2(x) − S2(−x) = 4 x (1−x2)2 = 4 x  S2(x2)   .

(14)

For |x| < 1 we could write expand this in a conventional Taylor series, but we can also expand this using (11) to get

F2(x) = ∞ ∑ n=1  4  n  x2n+1   ,   |x| < 1   .

(15)

Now rearrange (14) to write

U2(x) = 1 3 (F2(x) − S2(x)) = − 1 3 S2(−x) = − 1 3 (1+x)2   ,

(16)

where the reason for the factor of 1/3 will become obvious below.

Using the series expansions (11) and (15) we can write

U2(x) = ∞ ∑ n=1  n  cn(x)   ,   |x| < 1

(17)

where

cn(x) = 1 3 xn (4 xn+1 − 1)   .

(18)

It is obvious that

lim x → 1  cn(x) = 1   ,

(19)

for all n, so we can write

U₂(x) → 1 + 2 + 3 + 4 + … , x → 1^-     (20)

Where the 1⁻ indicates that x → 1 from below. However, from (16) we have

U2(1) = − 1 12   ,

(21)

so using this prescription

1 + 2 + 3 + 4 + …→ − 1 12   .

(22)

So what's going on here? In the limit x → 1⁻ the n^th term in the sum within (17) tends to the value n. The trick is that for any value x < 1 there are only a finite number of c_n(x) which are positive, and an infinite number that are negative. Take a look at the graph below, which shows c_n(x) for several values of x:

Once we get x > 0.9999 or so, a significant number of the c_n(x) are close to 1. However, for any x < 1, there are an infinite number of c_n(x) which are negative. These have weights much smaller than the postive values, but there are a lot more of them. The net result is that the sum (17) is always negative, and tends to −1/12 as x → 1⁻.

Is this a unique result? That is, do we always get

1 + 2 + 3 + 4 + …→ – 1 12

(23)

if we do it "properly"? Beats me. I would think that you could chose different a different set of functions c_n(x) which still have the property c_n(x) → 1 as x → 1, place them in (17), and get something other than -1/12 as x →1. But I haven't investigated that, and I certainly couldn't prove a general case.

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File translated with help from T_EX by T_TH, version 4.03.
On 1 Apr 2014, 18:17.