In previous
blog entries I have referred to this latter expression (which has its roots in
the L1 function) as the Alt L2 function.

Now with
respect to the Riemann zeta function, a special case occurs where s = 1,
yielding the harmonic series,

1/1 +
1/2 + 1/3 + 1/4 + …

This can
then be expressed in terms of combinations as

1/1C

_{0}+ 1/2C_{1}+ 1/3C_{2 }+ 1/4C_{3 }+ …
Now, with
respect to the denominator, we can define a new series where the nth term
represents the sum of the n terms of the previous series.

So we
thereby obtain

1/1 + 1/(1
+ 2) + 1/(1 + 2 + 3) + 1/(1 + 2 + 3 + 4) + …

= 1 + 1/3 +
1/6 + 1/10 + …

Now this
can equally be expressed in terms of combinations,

i.e. 1/2C

_{0}+ 1/3C_{1}+ 1/4C_{2 }+ 1/5C_{3 }+ …
So the get
this new series, we only need to increase the starting number of each
combination (in relation to the previous harmonic series) by 1.

Now if we
look at this later combination expression (which starts with 1/2C

_{0}), we find that its value is equal to the corresponding L2 function i.e. 1 + x^{1}+ x^{2 }+ x^{3 }+…, where x = 1/2.
So 1 + 1/2 +
1/2

^{2 }+ 1/2^{3 }+ … = 2, and^{ }
1 + 1/3 +
1/6 + 1/10 + … = 2.

And this is
universally the case for any integer k (k > 1, where x = 1/k) .

So when x =
1/k, then the L2 function i.e.

1/(1 –
1/k) = 1 + 1/k + 1/k

^{2 }+ 1/k^{3 }+ … has a corresponding Alt L2 expression as,
1/kC

_{0}+ 1/{(k + 1)C_{1}} + 1/{(k + 2)C_{2}} + 1/{(k + 3)C_{3}}_{ }+ …
Therefore
when for example k = 4,

1/(1 –
1/4) = 1 + 1/4 + 1/4

^{2 }+ 1/4^{3 }+ … = 4/3.
So this
represents the appropriate L2 function in this case where x = 1/2

^{2}= 1/4.
And this
can be given an Alt L2 expression as,

1/4C

_{0}+ 1/5C_{1}+ 1/6C_{2}+ 1/7C_{3 }+ …
= 1 + 1/5 +
1/15 + 1/35 + … = 4/3

In fact a
fascinating product type expression can also be given for this Alt L2 function.

So when as
in this case the starting value of the combination is 4, we let the numerator
and denominator of the 1st term = 4 * 3 * 2 * 1.

Thus the 1

^{st}term = (4 * 3 * 2 * 1)/(4 * 3 * 2 * 1) = 1.
Then to get
the next term we simply add 1 to each value in the denominator while holding
the numerator fixed.

So the 2

^{nd}term = (4 * 3 * 2 * 1)/(5 * 4 * 3 * 2) = 1/5.
And for
each subsequent term, we keep repeating by adding 1 to each previous value in
the denominator (while holding the numerator fixed).

So the 3

^{rd}term = (4 * 3 * 2 * 1)/(6 * 5 * 4 * 3) = 1/15,
and the 4

^{th}term = (4 * 3 * 2 * 1)/(7 * 6 * 5 * 4) = 1/35 and so on.
And the
existence here of 4 product terms (in relation to numerator and denominator)
coincides with the fact that the value of k (with respect to the L2 function) =
4.

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