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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 $\sum 1 = 0$.

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    $\begingroup$ Well, in order to derive a false statement, first assume $0=1$... $\endgroup$
    – Asaf Karagila
    Jun 28 '15 at 17:29
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    $\begingroup$ The statement is certainly false under the normal usage of $=$, which is also the only situation where your idea of an elementary derivation makes sense. Your usage of $=$ in the opening statement means "zeta function regularization returns this result" and nothing more. $\endgroup$ Jun 28 '15 at 17:40
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    $\begingroup$ That is not what the Wikipedia article cited says. That "equation" is not in the article. $\endgroup$
    – GFauxPas
    Jun 28 '15 at 17:40
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    $\begingroup$ @Renato: Unfortunately, as stated it is. I understand that this means "using zeta regulation to re-interpret the sum" or "some other meaning of convergence". But as stated, $\sum_{n\in\Bbb N} 1\neq-\frac12$. john mangual: (1) Wikipedia is wrong sometimes; (2) so what?; and (3) it is still false. $\endgroup$
    – Asaf Karagila
    Jun 28 '15 at 17:42
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    $\begingroup$ The part of the Wikipedia article you mention is at best ambiguous, if not plain wrong. The point here is that even if $\sum_{n = 1}^{\infty} \frac{1}{n^s}$ is divergent for some values of $s$, in particular for $s = 0$, there still is a unique analytic function $\zeta$ on $\Bbb{C}$ that assumes the value $\sum_{n = 1}^{\infty} \frac{1}{n^s}$ whenever this is well defined. In some contexts it may be useful to assume by convention that $\sum_{n = 1}^{\infty} 1 = \zeta(0)$, but that is entirely arbitrary and there is no elementary way to justify this. $\endgroup$
    – A.P.
    Jun 28 '15 at 17:57
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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.

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    $\begingroup$ I especially like the historical comments in your post. +1 $\endgroup$
    – Joel
    Jun 28 '15 at 18:33
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    $\begingroup$ You're addressing the alternating series $1-1+1-1+\cdots$, but the OP asked about the positive series $1+1+1+1+\cdots$. The latter is not Cesaro or Abel summable. $\endgroup$ Jun 30 '15 at 6:49
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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:

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

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  • $\begingroup$ I was writing my answer for too long and now it looks like I'm copying you. You have a $+1$ from me :) $\endgroup$
    – krvolok
    Jun 28 '15 at 18:30
  • $\begingroup$ @krvolok, that happens to me all the time. It's good to have several perspectives. If they take the same form, then that just means it's a good idea :) $\endgroup$
    – Joel
    Jun 28 '15 at 18:32
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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 :)

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    $\begingroup$ I doubt that Ramanujan summation can count as "an elementary derivation". $\endgroup$ Jun 28 '15 at 17:53
  • $\begingroup$ Could you prove Ramanujan summation as Euler-Maclaurin to 1st order? $\endgroup$
    – cactus314
    Jun 28 '15 at 17:54
  • $\begingroup$ @johnmangual Wikipedia does :) $\endgroup$ Jun 28 '15 at 18:00
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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 $~1+1+1+1+\ldots=(1-1+1-1+\ldots)+(0+2+0+2+\ldots)~=>~S=S_{\large\pm}+2S$

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


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 $\zeta(0)\equiv-\dfrac12$

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  • $\begingroup$ hmm? that seems plausible that $\sum (0+2) = \sum (1+1)$ and yet $\sum (0+2) = 2 \sum 1$. $\endgroup$
    – cactus314
    Jun 28 '15 at 17:52
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    $\begingroup$ This derivation allows you to have $S_\pm$ equal any number you wish. $\endgroup$
    – GFauxPas
    Jun 28 '15 at 17:53
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    $\begingroup$ @GFauxPas: Couldn't agree more... The slightly more intuitive explanation is that $\dfrac12$ is the mean between the two extremes, $0$ and $1$, which the partial sums alternately take. $\endgroup$
    – Lucian
    Jun 28 '15 at 18:03
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    $\begingroup$ Euler did not use this, he used what is now called Abel summation, using $1/(1+x)=1-x+x^2-\dots$ and it's derivatives, and passing to the limit $\lim\limits_{x\rightarrow 1^-}$ "carelessly". $\endgroup$
    – krvolok
    Jun 28 '15 at 18:05
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    $\begingroup$ You can prove anything if you glue together a lot of nonsense. $\endgroup$ Dec 18 '15 at 14:53
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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.

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    $\begingroup$ Where can one read about Ramanujan classes and all the interesting stuff you used in your answer? $\endgroup$ Feb 4 '16 at 14:00
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    $\begingroup$ What is a "Ramanujan class"? No find via google. $\endgroup$ Feb 17 '16 at 14:54

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