So for example when s = .5, ζ(.5) = 1/1^{1/2 }+ 1/2^{1/2 }+ 1/3^{1/2 }+ 1/4^{1/2 }+ ......
Therefore as the size of each term successively increases, the series diverges to infinity from this standpoint.
^{} However by analytic continuation the function can be given a finite value for all values of s between 0 and 1 (including 0).
^{} So it is very interesting in these circumstances that the extended Fibonacci sequence can be used very simply to provide approximations for all these values (between 0 and 1).
^{} The ratio in this case does not even require the generation of successive terms in the sequence, for values of s (between 0 and 1) correspond with the 1step case.
Therefore to compute the approximation  say  of ζ(.5), all the terms in the 1step sequence will increase at a constant rate of .5 i.e. 1, .5, .25, .125,.... Therefore the ratio stays fixed at .5.
So using the formula 1/(r_{s }– 1), we obtain the approximation 1/(.5 – 1) = 1/(– .5) = – 2.
This compares with the true value for ζ(.5) = – 1.460...
Therefore, though our approximation does indeed include a negative sign, it is still not very accurate in percentage terms.
However there is a simple way to improve such accuracy.
At the one extreme of the interval between 0 and 1, the extended Fibonacci approximation is 100% accurate i.e. where s = 1. Here the approximation, which is ∞, concurs exactly with the corresponding value of
ζ(1).
Then at the other extreme i.e. where s = 0, the extended Fibonacci approximation is 200% of correct value
For when s = 0, 1/(r_{n }– 1)  1/(0 – 1 ) = 1/– 1 = – 1.
However the true value for ζ(.5) = – .5.
So what is required is to put in a correcting factor which varies over the interval from 1 to 0 by 2^{k}
where k varies (in reverse fashion) from 0 to 1.
So we simply divide the estimate  obtained by the extended Fibonacci approximation  by 2^{k} .
Therefore the estimate for ζ(.5) = – 2/(2^{.5}) = – 1.414..
So this estimate does now indeed compare favourably with the true estimate of – 1.460.
And to illustrate further, ζ(.3) is now approximated by {1/((.3 _{ }– 1)}/2^{.7} = .879388
And even simpler way of obtaining these estimates for ζ(s), where s ranges from 0 to 1, can now be suggested.
This is given by the formula (s + 1)/2(s – 1)
So again when s = .5, we obtain (.5 + 1)/2(.5 – 1) = 1.5/(2( – .5) = – 1.5.
This again compares very favourably with the true value for ζ(.5) = – 1.460.
In the table below I give the actual values for ζ(s) from .1 to .9. I then give the approximations using the two methods (i.e. extended Fibonacci and simple formula). Finally I give the % accuracy for both approximations.
s

ζ(s)

Ex. Fibonacci
approx.

% Relative accuracy

Simple Formula approx.

% Relative
accuracy

0

– .500000

– .500000

100.0

– .500000

100.0

.1

– 603037

– .595429

98.7

– .611111

97.4

. 2

– .733920

– .717936

97.8

– .750000

97.9

.3

– .904559

– .879388

97.2

– .928571

97.4

.4

– 1.134797

– 1.099589

96.9

– 1.166666

97.3

.5

– 1.460354

– 1.414213

96.8

– 1.500000

97.3

.6

– 1.952661

– 1.894645

97.0

– 2.000000

97.6

.7

– 2.778388

– 2.707507

97.4

– 2.833333

98.1

.8

– 4.437538

– 4.352752

98.1

– 4.500000

98.6

.9

– 9.430114

– 9.330329

98.9

– 9.500000

99.2

So we can see that both approximations provide estimates for ζ(s) with a consistently high degree of accuracy over the interval from 0 to 1.
In fact since the first estimate consistently undershoots the correct value, while the second consistently overshoots it, a much more accurate estimate can be obtained by obtaining the average of the two estimates obtained.
Therefore for example for ζ(.5), our new estimate = (– 1.414213 – 1.500000)/2 = – 1.457106, which is now 99.8% accurate!
And again, we are not confined just to these estimates for 0 to .9. All values within the range of 0 and 1 can be calculated using the same basic formulae.
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