How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization

On Wikipedia we find $$\displaystyle \bbox[5px,border:1px solid #F5A029]{1 + 1 + 1+\dots =\sum_{n=0}^\infty 1 = -\frac{1}{2}}$$ using (the rather complicated) zeta-function regularization. I asking for an elementary derivation possibly based on the idea that on average

$$ \sum_{n=0}^\infty (-1)^n = \begin{cases} 1 & \text{ if }n\text{ is odd} \\ 0 & \text{ if }n\text{ is even} \\ \end{cases} = \frac{1}{2} $$

There is a blog discussion that uses very general "cutoff" functions, but I am having a hard time specializing to the case at hand.

A serious question is to qualify where the basic properties of addition break down:

$(a+b)+c = a+(b+c)$ associativity
$a+b = b+a$ commutativity
$a + 0 = a = 0 + a$ addition by zero

This example obviously shows we can't use these axioms infinitely many times without generating contradictions. A related question even has $\sum 1 = 0$.

Solution 1:

Of course, this series diverges, if we define summation of a series, $$\sum\limits_{n=0}^{\infty} a_n,$$in the usual way, as the limit of the sequence of partial sums, $\{s_n\}_{n\in\mathbb N}$, where $s_n:=\sum\limits_{i=0}^{n}a_i$.

However, a unique value can be assigned to some divergent series, if we agree to alter the definition of the summation itself (this is the "catch", of course) - we can redefine what the symbol $\sum_{n=0}^{\infty}$ means in order to broaden it. The question shifts, from what is $1-1+1-\dots$, to how shall we define $1-1+1-\dots$. Instead of the limit of sums, we define a summation method as a linear functional $S$ from sequences to $\mathbb R$. Likewise, we demand that we recover the same result if the partial sums converge in the ordinary sense, ie $$\lim\limits_{n\rightarrow\infty}s_n=s \:\Rightarrow S[a_n]=s.$$

Some summation methods also obey an additional property of stability, that if $a_0+a_1+\dots=s$, then $a_1+a_2+\dots=s-a_0$. I am restoring plus signs for easier understanding.

An example is given by the Cesaro $(C,1)$ sum, defined as $$S[a_n]=\lim\limits_{n\rightarrow\infty}\frac{s_0+s_1+\dots+s_n}{n+1}$$

which was actually first used by Leibniz to sum $1-1+1-\dots$ in 1713. It satisfies the third property as well.

An elementary derivation of the result you wish to know is given by Abel summation, $$S[a_n]=\lim\limits_{x\rightarrow 1^-} f(x)$$ where $$f(x):=\sum\limits_{n=0}^{\infty} a_n x^n$$ converges for $0\leq x < 1$. This was the method Euler used to sum $1+2+3+\dots$, and it reduces to Cesaro summation (it is slightly more general).

You can Abel sum your series, if you start from the geometric sum $\frac{1}{1+x}=1-x+x^2-\dots$, and let $x\rightarrow 1$.

For further discussion I recommend reading Divergent series by G. H. Hardy.

Solution 2:

There are several notions that generalize the summation of a series. One method is by Ramanujan, which you indicated you have seen, through the analytic continuation of the zeta function. Unfortunately, the two main methods will not work in this case, those being summation by Cesaro means and Abel summation. I will present them here in any case. This wont answer your question, but you may find the information useful.

One notion of summability through average is that of Cauchy. A series is said to be Cauchy convergent if the successive averages of the partial sums converge: $$\lim_{N\to \infty} \frac{S_1 + S_2 + \cdots + S_N}{N}.$$ If a series is convergent in the ordinary sense, then this limit converges to the sum. Unfortunately, this doesn't work out very well for the series you presented: $$\lim_{N\to \infty} \frac{ S_1 + S_2 + \cdots + S_N}{N} = \lim_{N\to \infty} \frac{ 1 + 2 + \cdots + N}{N} = \lim_{N \to \infty} \frac{N(N+1)}{2N} = \lim_{N\to \infty} \frac{N+1}{2} = \infty.$$

Another notion is that of Abel summations, where the terms of the series become the coefficient of a power series and the limit as $x \to 1^-$ is taken as the sum. Abel summation agrees with a Cauchy sum when the Cauchy sum exists, and thus it generalizes both Cauchy and ordinary convergence.

Thus if $\sum a_n$ is to be summed in the abel sense we consider: $$\lim_{x\to 1^-} \sum_{n=1}^\infty a_n x^n$$

In this case we have $$\lim_{x\to 1^{-}}\sum_{n=1}^\infty x^n = \lim_{x\to 1^-} \sum_{n=1}^\infty x^n = \lim_{x\to 1^-} \frac{x}{1-x} = \infty$$ so this series is not Abel summable either.

These are the two main approaches of summing a divergent series. You can find more information in Hardy's book on Divergent Series (a classic on the subject).

If you were to sum the divergent series $\sum_{n=0}^\infty (-1)^n$ in the Cesaro or Abel sense, this is how it would go down:

$$\lim_{N\to\infty} \frac{S_0 + S_1 + \cdots + S_N}{N} = \lim_{N\to \infty} \frac{1 + 0 + 1 + \cdots (1 \text{ or } 0)}{N} = \lim_{n\to\infty} \frac{ [N/2] }{N} = 1/2$$

And:

$$\lim_{x\to 1^-} \sum_{n=0}^\infty (-1)^n x^n = \lim_{x\to 1^-} \frac{1}{1+x} = \frac12$$

Solution 3:

You can use Ramanujan summation which for a function with no divergence at $0$ is defined as:

$$C(a)=\int_0^a f(t)dt-\frac 12 f(0)-\sum_{k=1}^\infty \frac {B_2k}{(2k)!}f^{2k-1}(0)$$

Now we choose C(0) as the result of the sum $\sum_{k=1}^\infty 1$ so the first integral vanishes like all the derivative of the function $f(x)=1$ so the sum which appears in the Ramanujan sum is 0, now we have only the term $-\frac 12 f(1)$ which is obviously $-\frac 12$.

This is a very powerful method to sum divergent sum :)