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

Question

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.

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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 $\sum 1 = 0$.

Answer 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.

Answer 2

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 :)

Answer 3

Euler did it like this: $~1-1+1-1+\ldots=1-(1-1+1-\ldots)\quad=>\quad S_{\large\pm}=1-S_{\large\pm}$

$=>\quad2S_{\large\pm}=1\quad=>\quad S_{\large\pm}=\dfrac12$

Then $~1+1+1+1+\ldots=(1-1+1-1+\ldots)+(0+2+0+2+\ldots)~=>~S=S_{\large\pm}+2S$

$=>\quad S=-S_{\large\pm}=-\dfrac12$

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Alternately, given that $\eta(k)=\big(1-2^{1-k}\big)~\zeta(k)$, since he already “established” that $\eta(0)\equiv\dfrac12$ ,

it follows that $\zeta(0)\equiv-\dfrac12$

Answer 4

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:

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

Answer 5

For $f(x)=\sum_0^\infty a_n x^n$, a real number $R$, $R\neq 1$, and $g(x)=f(x)-Rf(x^2)$, if $g(1)$ is Abel summable, that is, $g(1)=\lim_{x\to 1^-} g(x)$, an elementary Ramanujan sum of $f(1)$ is defined by $$ f(1)=\frac{g(1)}{1-R}. $$ Inversely, for an Abel summable series $g(1)=\sum_0^\infty g_n$ and a real number $R$, $R\neq 1$, the corresponding elementary Ramanujan summable series $f(1)$ is defined by $$ a_0=\frac{g_0}{1-R}~,~~a_{2n+1}=g_{2n+1}~,~~a_{2(n+1)}=g_{2(n+1)}+Ra_{n+1}~. $$

$f(1)$ is an element of the linear space of elementary Ramanujan summable series corresponding to $R$, the Ramanujan class $R$. If $f(1)$ is an element of two distinct classes, $f(1)$ is Abel summable, that is, an element of the class $R=0$.

One should not naively sum series of distinct classes (here).

For $F_k(x)=f(x^k)$, $F_k(1)=f(1)$ since $\lim_{x\to 1^-} g(x^k)=g(1)$. $F_k(1)=a_0+0+\cdots +0+a_1+0+\cdots +0+a_2+\cdots$, with $k-1$ zeros between $a_n$ and $a_{n+1}$. $F_k(1)$ is an element of the class $R$, as $f(1)$.

For $f'(x)=x\frac{df(x)}{dx}$, $f'(1)$ is an element of the class $2R$.

Consider $f(x)=\frac{x}{1-x}=x+x^2+x^3+\cdots$, $g(x)=\frac{x}{1+x}=x-x^2+x^3-\cdots$ and $R=2$. $g(1)$ is Abel summable to 1/2 and the elementary sum of $f(1)$ is -1/2.

$f'(1)=1+2+3+\cdots = -1/12$ is an element of the class $R=4$ and $f''(1)=1+4+9+\cdots = 0$ is an element of the class $R=8$ etc.

Consider now $f(x)=x+2x^2+4x^4+\cdots$, $g(x)=x$ and $R=2$. The elementary sum of $f(1)=\sum_1^\infty a_n$, with $a_n=n$ if $n$ is a power of 2 and $a_n=0$ if not, is $-1$.

$f'(1)=\sum_1^\infty a_n^2=-1/3$, $f''(1)=\sum_1^\infty a_n^3=-1/7$ etc.

To verify that $1-1+2-2+3-3+\cdots=1/8$, consider $$ f(x)=x-x^2+2x^3-2x^4+3x^5-3x^6+\cdots=\frac{1}{(1+x)^2}\frac{x}{1-x}~, $$ $$ g(x)=\frac{x-x^2-x^3-x^4}{(1+x)^2(1+x^2)^2} $$ and $R=2$.

Edit. The consistency of (iterated) elementary Ramanujan summation with analytic continuation of Dirichlet series is shown here (this question is deleted!), here (this question is deleted!) and here.

Answer 6

The summation from zero to infinity is plus one half, not minus one half. The zeta function is defined as a summation that starts at $n = 1$, and this yields minus one half, therefore the summation that starts at zero should yield plus one half.

Another way to derive this is via the identity:

$$\sum_{k=0}^{\infty}f(k) = \int_0^{\infty} f(x) dx + \int_{-1}^0 S(x) dx$$

where $S(x)$ is the analytic continuation of the partial sum $S(n) = \sum_{k=0}^n f(k)$. This is very simple to prove for convergent series. You can then start with assuming that $S(x)$ is a given function and then you have $f(x) = S(x) - S(x-1)$. Putting $x = 0$ in here and using that $S(0) = f(0)$ then implies that $S(-1) = 0$.

The summation is the limit of $S(x)$ for $x$ to infinity and this should then be equal to the limit of $n$ to infinity of the integral from $n-1$ to $n $ of $S(x)$. The right hand side is a telescoping expression for that.

In case the summation is divergent, the integral on the right hand side should be regularized by cutting it off at an upper limit of $R$ and then omitting all positive powers of $R$ and then taking the limit of $R$ to infinity. Here we appeal to the existence of an analytic continuation without being explicit about it. The idea is then that the function we're summing could be deformed using one or more parameters, and that for different values of those parameters than the one that yields the function that we want to sum, these positive powers would be negative powers, and the integral and summation would converge.

So, for those other values of these hidden parameters, the limit of $R$ to infinity exists. If we then analytically continue the results from there to here, the terms which diverge here are killed there so they should be dead on arrival here.

In this case we have $f(x) = 1$, so $\int_0^R f(x) dx = R$, and we should then disregard this term. We have $S(n) = n+1$, so for all real $x$ we have $S(x) = x+1$. We then find: $$\sum_{k=0}^{\infty}1 = \int_{-1}^0 (x+1)dx = \frac{1}{2}$$

Yet another method is to use the formula:

$$\int_0^{\infty}\frac{g(x)}{1+x}dx = -\sum_{k=0}^{\infty}\frac{dc_k}{dk}\tag{1}$$

where the $c_k$ are given by the series expansion coefficients of $g(x)$:

$$g(x) = \sum_{k=0}^{\infty}(-1)^k c_k x^k$$

This is based on Ramanujan's master theorem, and I've explained the derivation of this result in detail here. We can then use this formula to sum series by integrating minus the summand to compute $c_k$, and then we compute the alternating generating function $g(x)$, and then the summation follows from (1).

We should then note that in case of a diverging series, if we obtain a convergent integral expression this way, then for that integral (1) won't be valid, because if it were valid, the summation on the r.h.s. would have to be a convergent one. The argument hat the integral we obtain this way yields the correct regularized sum is then again an analytic continuation argument. We imagine that the function we are summing can be deformed using one or more parameters, and that there exists a domain for these parameters where the summation is convergent and where the procedure to get to (1) yields a valid identity for the summation.

If the integral in (1) is then also an analytic function of the parameters then analytically continuing the result from the domain where the summation converges to the values that yield the diverging summation should then yield the value of the integral that we get by plugging in $g(x)$ obtained directly from the diverging series, if that yields a convergent integral.

So, in this case we have:

$$-\frac{dc_k}{dk} = 1$$

therefore:

$$c_k = k$$

The integration constant is chosen to be zero, as that yields a converging integral. We can then comute the alternating generating function using the geometric series:

$$\frac{1}{1+x} = \sum_{k=0}^{\infty}x^k$$

Differentiating this yields:

$$-\frac{1}{(1+x)^2} = \sum_{k=0}^{\infty}k x^{k-1}$$

Multiplying both sides by minus $x$ yields:

$$\frac{x}{(1+x)^2} = \sum_{k=0}^{\infty}k x^k$$

So, we need to put:

$$g(x) = \frac{x}{(1+x)^2} $$

in (1). This yields the result for the summation as:

$$\int_0^{\infty}\frac{x}{(1+x)^3}dx = \int_1^{\infty}\frac{t-1}{t^3}dt = \frac{1}{2}$$