What is the sum of all the Fibonacci numbers from 1 to infinity. 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 $F = -3$.

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)
