Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? [duplicate]

I just found the sum of all Fibonacci numbers and I don't know if its right or not.

The Fibonacci sequence goes like this : $1,1,2,3,5,8,13,\dots$ and so on

So the Fibonacci series is this $1+1+2+3+5+8+13+\dots$

Let $1+1+2+3+5+8+\dots=x$

\begin{align} 1 + 1 + 2 + 3 + 5 + \dots &= x\\ 1 + 1 + 2 + 3 + \dots &= x\\ 1 + 2 + 3 + 5 + 8 + \dots &= 2x \text{ (shifting and adding)} \end{align}

We in fact get the same sequence. But the new sequence is one less than the original sequence. So the new sequence is $x-1$. But $x-1=2x$ which implies that $x=-1$.

So $1+1+2+3+5+8+\dots=x$ which means... $1+1+2+3+5+8+13+21+\dots=-1$

Is this right or wrong? Can someone please tell? Thanks...

• This is called divergent sum renormalisation. See Ramanujan's method. Commented Jan 4, 2018 at 8:37
• You have come to the correct conclusion in that divergent sum renormalisation techniques do indeed assign the value $-1$ to this series. Commented Jan 4, 2018 at 8:43
• A book on this subject would be Hardy - Divergent Series. Commented Jan 4, 2018 at 10:17
• The sum doesn't exist. The series diverges, as pointed out several times by others. However, there are ways of assigning real numbers to series other than summing them. Some of these methods even coincides with the sum whenever the sum exists, and thus bears some resemblance to the notion of sum that we know. You have found one such method, and that method assigns the value $-1$ to the Fibonacci series. It would also assign $-\frac1{12}$ to the series of the natural numbers, which gives users on this site headaches whenever it pops up. Commented Jan 4, 2018 at 10:32
• @Arthur and yet that -1/12 result seems to have practical applications on quantum physics... (i do get the same headaches too). Commented Jan 4, 2018 at 15:07

You're assuming that the limit of the sum of the first $n$ Fibonacci numbers exists as $n \to \infty$, which it doesn't. Which is to say that in order to apply your method, the series must be convergent but the series diverges, so your method is wrong. It is, however, related to the power series of the Fibonacci numbers, that is to say $$1+z+2z^2+3z^3+5z^4+8z^5+... = \sum_{n=0}^\infty F_{n+1}z^n=\frac{1}{1-z-z^2}$$

One can see that when we put $z=1$, the value is $-1$, as you assert in your answer.

However, this series is only convergent when $|z|<\dfrac{1}{\phi}$, where $\phi$ is the golden ratio, as should follow from Binet's Formula. See here for a derivation of the upper identity, and more information.

The problem is that the series you're trying to sum is divergent. You cannot manipulate divergent series by rules you can use with absolutely convergent series! Otherwise, by following the same "method" as yours, I can also claim the ridiculous statement that $1+2+3+\cdots=0$ as follows: \begin{align} 1 + 2 + 3 + 4 + 5 + \dots &= x\ \fbox1\\ 1 + 2 + 3 + 4 + \dots &= x\\ 1 + 3 + 5 + 7 + 9 + \dots &= 2x\ \fbox2 \end{align} Also multiplying $\fbox1$ by 2 gives $$2+4+6+8+10+\dots=2x\ \fbox3$$ So $\fbox1=\fbox2+\fbox3$ $\implies$ $x=2x+2x=4x$ $\implies$ $x=0$ as $1\ne4$.

Obviously this is totally absurd. This is because I'm manipulating a divergent series by rules only allowed to be used with absolutely convergent series. If a series is divergent, it diverges – trying to make anything out of it is just pointless.

• "If a series is divergent, it diverges – trying to make anything out of it is just pointless" google eulers golden nugget. Commented Jan 4, 2018 at 9:05
• @Arjang Unfortunately, this does not prevent that many people keep on claiming that it is valid in some sense. Commented Jan 4, 2018 at 18:43
• @Peter : of course it is valid in many senses : youtube.com/watch?v=XFDM1ip5HdU Commented Jan 4, 2018 at 19:56

Since the terms of the series don't go to $0$, the series does not converge. Relatedly, $x=\infty$ also solves your equation $x-1=2x$.

• Good point, although $\infty$ should not be treated as a number because doing this can lead to absurd consequences. IF the series WOULD converge, the limit WOULD be $-1$, but since it doesn't, the limit isn't $-1$, so the equation is simply false. Commented Jan 4, 2018 at 18:54
• Peter, I agree, thanks! Commented Jan 4, 2018 at 20:38

This kind or reasoning (let $\sum{a_n}=x$ and so on) is valid iff the series converges. Otherwise is just a mathematical formalism, nice, but meaningless (at least in these context. Probably using Analytic continuation you can give some motivation to your results using generating functions). Obviously the fibonacci's succession doesn't converge to $0$, and so the series cannot converge.

• "Probably using Cesàro summation you can give some motivation to your results." As the wikipedia article you link to indicates, if the terms of a sequence diverge to infinity, it is not Cesaro summable. Commented Jan 4, 2018 at 8:53
• @PeteL.Clark You're right, thanks for the hint
– user515010
Commented Jan 4, 2018 at 9:07
• @gabrielecassese I agree to "meaningless", I do not know what should be "nice" , but that is a matter of taste. At least many fake proofs can be made this way, for those who like such fake proofs. Commented Jan 4, 2018 at 18:46