Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? 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...
 A: 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.
A: Since the terms of the series don't go to $0$, the series does not converge. Relatedly, $x=\infty$ also solves your equation $x-1=2x$. 
A: 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.
A: 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. 
