# [seqfan] Re: Fibonacci-sums

Gottfried Helms Annette.Warlich at t-online.de
Thu Aug 19 15:01:49 CEST 2010

```Am 19.08.2010 07:08 schrieb Robert Israel:
>
> Note that fib(k) = (phi^k - (-1/phi)^k)/sqrt(5), where
> phi = (sqrt(5)+1)/2.
> So formally
> sum_{k=0}^infty (-1)^k fib(k) = (1/(1 + phi) - 1/(1 - 1/phi))/sqrt(5)
> which simplifies to -1.

Well, can be done even simpler without the explicite squareroots
in the Binet-formula, just single-digit integer arithmetic... :-)

The matrix-expression to get the fibonacci-numbers is

m0 = [1,1]
[1,0]

then
m0^n = [ fib(n+1)  fib(n) ]
[ fib(n)    fib(n-1)]

Example

m0^3 = [ 3  2 ]
[ 2  1 ]

Now the alternating sum of all fibonacci-numbers should agree
to the alternating sum of all powers of m0.
This is the alternating geometric series, and its closed form
is in general 1/(1+x) and adapted to the matrix m0:

as = (I + m0)^-1

Indeed we get

as = [ 1  -1 ]
[-1   2 ]

and the expected sum (-1) at the positions, where the summation
starts with its first term fib(0) at m0^0.

I can even follow the same idea for the nonalternating sum:

ps = ( I - m0)^-1
to get
ps = [-1 -1 ]
[-1  0 ]

from which one could conclude, the non-alternating sum is (-1).

But I don't trust the latter computation. Nonalternating sums
which diverge have usually nasty specialities...

(Btw., the same would come out if I used diagonalization and the
geometric series of the elements of the diagonal, we have then the
geometric series of phi-related-values in the diagonal )

Gottfried Helms

```