∑{ζ(s)
– 1} ~ 1 (for s = 2, 3, 4,…..)
For example from adding up values for s = 2 to 10, we obtain
.6449 + .20205 + .08232 + .03692 + .0173 + .00834 + .00407 +
.002008 + .000904
= .99812 (which is already close to 1)
Then ∑{ζ(s) – 1}
~ .75 (for even values of s i.e. s =
2, 4, 6, …)
So for even values of s from 2 to 10, we obtain
.6449 + .08232 + .0173 + .00407 + .000904
= .749494 (which again is very close to .75).
Also ∑{ζ(s) – 1} ~ .25 (for
odd values of s i.e. 3, 5, 7, ….)
So for odd values of s from 3 to 10, we obtain
.20205 + .03692 + .00834 + .002008
= .249318 (which for just 4 values computed is again close
to .25)
There are also interesting connections as between the
Riemann zeta function (for positive integer values of s and the Euler-
Mascheroni constant i.e. γ = .5772156649…
As is well
known for ζ(s) where s = 1 (i.e. the harmonic series) and
the summation of the series is taken over a finite set of values n,
ζ(1) = log n + γ
However γ in turn
is related to all ζ(s) - now summed without limit - for the other positive integer values
of s in the following manner!
γ = ζ(2)/2 – ζ(3)/3 + ζ(4)/4 – ζ(5)/5 + ……
So for s =
2 to 10, we obtain
1.644934/2 – 1.202056/3 + 1.082323/4 – 1.03692/5 + 1.0173/6 – 1.00834/7 + 1/00407/8 –
1.002008/9 + 1.000904/10
= .62474….
Now this
approximation is still not very accurate, but in this case the series the
series diverges very slowly towards the true value (oscillating alternating
above and below the true value).
A better
approximation however can be obtained as follows:
1 – γ
= {ζ(2)/2 – 1}/2 + { ζ(3)/3 – 1}/3 + {ζ(4)/4 – 1}/4 + {ζ(5)/5 – 1}/5 + ……
So again
summing for s = 2 to 10, we obtain
.644934/2 +
.202056/3 + 082323/4 + .03692/5 + .0173/6 + .00834/7 + .00407/8 + .002008/9 +
.000904/10
= .42268
(correct to 5 decimal places) which gives γ = .57732 which is already a
very good approximation to the true value i.e. .5772156649…
Also ζ(s)/ζ(s + 1) ~ 1, and
{ζ(s) – 1}/{ζ(s + 1) – 1} ~ 2, again
for sufficiently large t.
For example ζ(9)
= 1.002008 and ζ(10) = 1.000904
Therefore ζ(9)/
ζ(10) = 1.002008/1.000904 = 1.0011… (which is already close to 1)
Likewise ζ(9)
– 1 = .002008 and ζ(10) – 1 = .000904
Therefore {ζ(9)
– 1}/{ζ(10) – 1} = .002008/.000904 = 2.2212…
This is not yet very close to 2. However for larger t the
ratio will progressively fall towards 2!
In all cases i.e. for positive integers > 1, ζ(s) can be expressed as 1 + k
(where k is less than 1)
So for
example ζ(2) = 1.6449… = 1 + .6449…
We can then
define a “complementary” number as 1 – k
So in the case of ζ(2), 1 – k = 1
– .6449… = .3551
We can now define a new set of twin relationship as πs/ts1 = 1 + k and πs/ts2 = 1 – k respectively.
ts1 and
ts2 new are
the two denominators associated with the common numerator πs.
For example
when s = 2, π2/6 = 1 + .6449… and π2/27.79… = 1 – .6449… respectively.
So here, ts1
= 6 and ts2 = 27.79… respectively
And the
difference of ts2 and ts1 = 27.79 – 6 = 21.79…
When s grows sufficiently large {ts2
– ts1}/{t(s + 1)2 – t(s + 1)1} ~ 2/π
For example when s = 9, k = .002008; ts1 = 29749 (to nearest integer) and ts2 = 29869.
Therefore ts2
– ts1 = 120.
When s =
10, k = .000904; t(s + 1)1 =
93555 and t(s + 1)2 = 93733
Therefore t(s
+ 1)2 – t(s + 1)1 = 178
So {ts2 – ts1}/{t(s + 1)2 – t(s
+ 1)1} = 120/178 = .6741…
This
compares fairly well with 2/π = .6366..
And the approximation steadily improves for larger s.