Today I believe I had found the sum of all the Fibonacci numbers are from $1$ to infinity, meaning I had found $F$ for the equation $F = 1+1+2+3+5+8+13+21+\cdots$ I believe the answer is $-3$, however, I don't know if I am correct or not. So I'm asking if someone can show how $F = -3$.


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    $\begingroup$ I'm curious...what exactly lead you to believe that the sum is $-3$? $\endgroup$ – MathematicsStudent1122 May 3 '16 at 3:05

None of these answers adhere to analytic continuation, which is clearly what you are looking for. A fairly non-rigorous method for doing this is to use the generating function for the Fibonacci series, namely $$\frac{1}{1-x-x^2} = 1 + 1x + 2x^2 + 3x^3 + 5x^4 + 8x^5 +\cdots$$ Clearly we get the answer $-1$ if we plug the number $1$ into each side. This does not give $-3$.... I am not sure how you would even get that. Again, this is VERY non-rigorous... see this post by Terence Tao for a better foundation into the topic. While it is true the sum diverges in the classical sense, it is often possible to give finite values to sums which have meanings in some contexts (and are simply absurd in others!). For a basic example, look up Cesaro summation, or even Abel summation. https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

Please see this link! Apparently a zeta regularization for the Fibonacci series has been done, and matches my answer.

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    $\begingroup$ You seem to use the words "clearly" and "rigid" in... interesting ways. $\endgroup$ – Rahul May 3 '16 at 3:08
  • $\begingroup$ @Rahul XD Yes, I suppose I use them here to refer to the fact that while it is easy to throw the number $1$ into both sides and get an answer, doing so goes outside of the interval of convergence and is thus frowned upon. The zeta regularization I mention at the end and link to is much more "rigid" in my use of the word, i.e. doesn't involve naively violating the laws of arithmetic to get some answer. If you have better term recommendations feel free to share :) $\endgroup$ – Brevan Ellefsen May 3 '16 at 3:11
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    $\begingroup$ I think you mean "rigorous" rather than "rigid." $\endgroup$ – Thomas Andrews May 3 '16 at 3:52
  • $\begingroup$ I think we can also interpret this $p$-adically: if we take $\epsilon>0$, then we can compute $\sum_{n\geq 0} F_n p^{n\epsilon} = \frac{1}{1-p^\epsilon - p^{2\epsilon}}$ in $\mathbb{C}_p$. Letting $\epsilon \to 0$ gives a limit of $-1$. I suspect that we can also do this as a $p$-adic Cesàro sum. $\endgroup$ – Slade May 3 '16 at 22:20

You are incorrect. The infinite sum of all the Fibonacci numbers diverges to infinity. Think of it this way: for positive integer I take, say $M$, I can always sum enough terms of the Fibonacci sequence to get a higher value than $M$, And, this works for any $M$ we choose. Therefore, the sum is larger than any positive integer, so it must diverge to infinity.

  • $\begingroup$ The sum of all natural numbers equals $-\frac{1}{12}$, a result used in some physics applications. There's a youtube video on it by numberphile, perhaps the OP is looking for a similar proof. $\endgroup$ – Ninosław Brzostowiecki May 3 '16 at 2:51
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    $\begingroup$ It was on Youtube, so it must be true. $\endgroup$ – user247608 May 3 '16 at 2:53
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    $\begingroup$ @MathematicsStudent1122 That "sum" is vastly different than normal addition. You can prove $1-1+1\dotso =1/2$ by Cesaro summation. But by standard addition and notions of convergence it diverges. Its good to know which summation method youre talking about. $\endgroup$ – Nap D. Lover May 3 '16 at 2:58
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    $\begingroup$ Not dismissing, perhaps having reservations. Claims like this are misinterpreted by most people and the claims have to be qualified by what the particular summation means. For most people, the result is not meaningful. $\endgroup$ – user247608 May 3 '16 at 3:09
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    $\begingroup$ I have never seen a justification that the sum of the natural numbers "is" $-1/12$. I have seen a great exposition by Terence Tao justifying that $-1/12$ is the constant term of a canonical asymptotic expansion of that series, but calling this the "value" seems highly situational. $\endgroup$ – Slade May 3 '16 at 23:32

All Fibonacci numbers are positive integers (and additions of positive integers), and addition of positive integers is closed under positive integers, thus $-3$ is wrong by default, even if it it didn't diverge (which it does).

Unless you are referring to simply assigning it a value similar to Ramanujan summation, then you are incorrect.


In particular to @Brevan Ellefsen.

This is not an answer, but it's more than a simple remark.

Have you ever met the paradoxal identity

$$-1=\sum_{k=0}^{\infty}2^k \ \ \ (1) \ ?$$

see http://www.cut-the-knot.org/blue/p-adicExpansion.shtml and many other references such as Knuth (exercice 21 p 213 of Vol. 2 of The Art of Computer programming) with explanation in considering the development of

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

in which $x=2$, thus beyond its (usual meaning) disk of convergence.

A rigorous framework (I don't say "the" framework) justifying this kind of result is the concept of p-adic number, or, in a preferable manner, of the so-called p-adic topology. For example, in the 2-adic topology, the "size" of an integer having a prime factor decomposition $2^{k_2}3^{k_3}5^{k_5}...$ is $2^{-k_2}$: the more factors $2$ a number possesses, the smallest it is... Thus, it is not so strange that the RHS of (1) converges. The fact that it is equal to $-1$ is not that much complicated to establish...

Let it be clear: I don't say I have a convergence proof for $\sum_{k=0}^{\infty}F_k=-1$ (or $-3$...) along the same line, for example by using the generating function $\dfrac{1}{1-x-x^2}$. I just say that, with a little time (that I have not for the moment) one could hopefuly obtain a result using this kind of topology.

Small introduction: p-adic topology is a very specific way to topologize $\mathbb{Z}$ (one topology for each prime $p$). $\mathbb{Z}$ with this structure is denoted $\mathbb{Z}_p$ (same notation as integers mod. p but no direct connection). In particular, it is an ultra metric topology, the "archimedean" property of integers is no longer available, which explains why some people look scandalized : they are not aware of the existence of other topologies giving other limits than $\infty$.

For readers that are very mathematically inclined, have a look at the famous book "Number Theory" of André Weil (one of the mathematicians that has contributed to a famous conjecture that has led to the final demonstration of Fermat's theorem by Andrew Wiles) which builds structures on $\mathbb{Z}_p$: $\mathbb{Q}_p$, adeles, ideles...

To have an idea of the deep mathematics that are involved in these questions, see

The Geometry of p-Adic Fractal Strings: A Comparative Survey (Lapidus and Hung) http://www.math.ucr.edu/~lapidus/papers/ContMath//GeometrypAdicStringsSurvey10893.pdf

A direct reference where the generating function of Fibonacci numbers is considered in the p-adic framework is


(some typographical problems with Japanese characters there)

For entering into the subject of p-adic topology, here is a nice article: https://www.colby.edu/math/faculty/Faculty_files/hollydir/Holly01.pdf.

For an elaborate theory:


See also the rather recent articles, the first one dealing with Fibonnacci numbers in a p-adic context:


http://arXiv:0809.0492v1 "From Data to the p-Adic or Ultrametric Model" by Fionn Murtagh

Remarks: 1) the Wikipedia article https://en.wikipedia.org/wiki/P-adic_numbers is well done, but its first part: the 10-adic numbers should not be part of the picture, because taking a non-prime $p$ makes a very poor structure.

2) I have met, in a "mathematics for electronics" book an "explanation" linking the "2's complement" hexadecimal notation "-1=FFFF" to identity (1)

  • $\begingroup$ Very interesting to read, and very informative! I briefly looked through some of the articles you shared... I hope to read more carefully later. Concerning that second remark... I remember reading that somewhere as well, but I can't quite recall where. I managed to find this, though I would be interested in finding a better "explanation" connecting the 2-adic numbers and 2's Complement. $\endgroup$ – Brevan Ellefsen May 4 '16 at 23:45

The sum diverges to infinity, but a Shanks transformation (https://en.wikipedia.org/wiki/Shanks_transformation) converges to -1.

In effect this Shanks transformation, too, corresponds to an analytical continuation of the Maclaurin series for $1/(1-x-x^2)$.


OK, now I will explain how I got -3 for the sum of all Fibonacci numbers from 1 to infinity. This may not make sense, but hopefully this will explain how I got -3.
I started thinking of a series where the first Fibonacci number was added by 1, the second Fibonacci number was added by 2, and so on. the first 8 numbers of this series that I had created were as follows: 2 (1+1), 3 (1+2), 5 (2+3), 7 (3+4), 10 (5+5), 14 (8+6), 20 (13+7), and finally 29 (21+8). To refer to the nth number in this sequence I will write $J(n)$, so $J(1) = 2$, $J(2) = 3$, etc. I soon noticed that if you added $J(n)$ and $J(n+1)$ you would get $J(n+2)+n-1$. I rearranged this so that it looked like this: $J(n)=J(n-2)+J(n-1)-n+3$. I used this to find $J(0)$ and $J(-1)$ (both were 0), but this will be used later.
I then started thinking of what J, the sum of all the numbers in the series from 1 to infinity, could be, but I then realized that, because each term of this series was a Fibonacci number + a natural number, $J=F-(1/12)$ where F is the sum of all the Fibonacci numbers from 1 to infinity, and -1/12 was the sum of all natural numbers. Because J also equals $J(1)+J(2)+J(3)+...$ I decided to apply the "formula" ($J(n)=J(n-2)+J(n-1)-n+3$) to each term to get $J=J(-1)+J(0)+2+J(0)+J(1)+1+...$ Rearranging this, I got $J=J(-1)+2(J(0)+J(1)+J(2)+J(3)+...)+a$ where $a=2+1-1-2-3-...$ Multiplying both side by -1, I found $-a=-2-1+1+2+3+4...$ which can also mean $-a=-3-(1/12)$ This lead me to find that $a=3+(1/12)$
Looking back at the J equation ($J=J(-1)+2(J(0)+J(1)+J(2)+J(3)+...)+a$) , inside the parenthesis next to the 2, I saw that it equaled $J+J(0)$, which is J since $J(0)=0$ and $J(-1)=0$. This equation then became $J=2J+3+(1/12)$, and by solving for J, I found That $J=-3-(1/12)$ (which is -a). Locking back at the $J=F-(1/12)$ equation, hopefully, you can see how $F=-3$ (And hopefully I am not crazy)

  • $\begingroup$ Your answer was a bit challenging to read through, but it does explain how you got to $-3$. Might I suggest breaking it up into a few paragraphs at least? $\endgroup$ – Corellian May 3 '16 at 22:29
  • $\begingroup$ @Brody I don't know how to make paragraphs, but I would add paragraphs if I knew. $\endgroup$ – Jessep May 3 '16 at 22:32
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    $\begingroup$ To make a paragraph break press spacebar twice and then enter. You do a lot of niave operations on the terms here, and, without looking too carefully through everything you wrote, I feel like that is the root of your problem. You could develop a summation system in which your answer makes sense, but your answer simply doesn't match the results from the currently accepted system (i.e. Ramanujan summation, Abel summation, Cesaro summation, Zeta regularization, etc.). For example, as noted in the link in my post, the sum $$2+3+4\dots = -7/12 \neq 0+2+3+4\dots = -\frac{13}{12}$$ $\endgroup$ – Brevan Ellefsen May 3 '16 at 23:08

F = 1 + 1 + 2 + 3 + 5 + 8 + 13 + ...

F = 0 + 1 + 1 + 2 + 3 + 5 + 8 + ...

2F = 1 + 2 + 3 + 5 + 8 + 13 + 21 + ...

2F = F-1

F = -1

  • $\begingroup$ This does not particularly add anything to existing answers. But it is interesting. $\endgroup$ – 6005 May 11 '17 at 15:25
  • $\begingroup$ Thanks. It is the easiest way I found to explain. $\endgroup$ – Haran May 13 '17 at 8:35

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