[seqfan] Fibonacci-sums

Gottfried Helms Annette.Warlich at t-online.de
Wed Aug 18 13:59:48 CEST 2010

```Reading the wikipedia-article on fibonacci-numbers
I tried what would be the alternating sum of
all fibonacci-numbers. The sequence is diverging,
so one has to employ a method of divergent summation.

Happily the growthrate is not too strong, so simple
Euler-summation suffices.

Using fib(0)=0,fib(1)=1, fib(2)=1, fib(3)=2,... I got

inf
sum    (-1)^k * fib(k)   = -1     // Eulersummation
k=0

Well, it is interesting, that it gives such a simple value.

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It calls for extension. I used powers of the index as cofactors

inf
sum   - (-1)^k * fib(k) * k^n   =  a01(n)     // Eulersummation
k=0

and got the following sequence a(n), n beginning at 0:

a01(0...5) = 1,2,8,50,416,4322

The sequence is known in OEIS:

*  A000557:    From Fibonacci sums.
*         1, 2, 8, 50, 416, 4322, 53888, 783890, 13031936, 243733442

The commentary gives a finite formula

and
*  a(n) = Sum[k=0..n, k!*Stirling2(n, k)*Fibonacci(k+2)].

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fib(0)=1,fib(1)=1, fib(2)=2, fib(3)=3,... this gives

a02(0..5) =   1, 1, 5, 31, 257, 2671,

Also this sequence is known to OEIS:

*   A000556:  Expansion of exp(-x) / [ 1 - exp(x) + exp(-x) ].
*         1, 1, 5, 31, 257, 2671, 33305, 484471, 8054177, 150635551,

The commentary gives a computation by finite sums:

*   FORMULA:   Sum(k!*fibonacci(k + 1)*stirling2(n, k), k = 0 .. n).

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Using binomials as cofactors I get the fibonacci-numbers again:

inf
sum   - (-1)^k * fib(k) * binomial(k,n)   =  a03(n)     // Eulersummation
k=0

a03(0..5) = 1,2,3,5,8,13

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Hmm. It would be nice to see, whether we can define values for the
infinite non-alternating sums....

With cofactors 1/n^k , abs(n) > 1    I get

inf                                    n
sum    fib(k) /n^k   =  a04(n)   = -----------
k=0                                 n^2 - n -1

whose denominator looks familiar in context with fibonaccimumbers.
It seems valid also for fractional/real values of n. Also this
agrees with formula a01, if I insert n=-1 . So possibly I can
assume, that analytical continuation will allow also to insert
n=1 and to get the result a04(1) = -1 .
But how could this be proved/dealt with?

Gottfried Helms

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Pari/GP:
{  fibseq(k,f0=0) = if(k==0
,n1=f0;n2=1;n3=n1+n2
,n1=n2;n2=n3;n3=n1+n2);
return(n1) }

a01(n) = - sumalt(k=0,(-1)^k* fibseq(k,0) * k^n )
a02(n) =   sumalt(k=0,(-1)^k* fibseq(k,1) * k^n )
a03(n) = - sumalt(k=0,(-1)^k* fibseq(k,0) * binomial(k,n) )

========================================================================

```