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...

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.
