## Monday, April 25, 2016

### Surprising Result

As is well known the sum of the harmonic function,

1 + 1/2 + 1/3 + 1/4 + 1/5 + ............ ~ log n + γ (where γ = .5772... represents the Euler-Mascheroni constant).

It is also known (from Dirichlet) that the average no. of divisors (natural number factors) of n

~ log n – 1 + 2γ.

It then struck me how the second result could be related to the first.

In the numbers up to n every number must necessarily include 1 as a factor, then every second number must include 2 as a factor, every 3rd number, 3 as a factor, every fourth number, 4 as a factor and so until finally every nth number includes n as a factor.

Therefore if we were to calculate the average no. of factors to n, this would imply that it would equate to the sum of

1 + 1/2 + 1/3 + 1/4 + ...... + 1/n.

However there is a slight problem with this in that the divisors must in practice be whole numbers.

Therefore when we use the harmonic series, it also includes remainders, thereby over-estimating our result.

For example if we were to consider the total no. of factors to n = 10, using the harmonic series, we would obtain

10 + 5 + 3.33 (instead of 3) + 2.5 (instead of 2) + 2 + 1.67 (instead of 1) + 1.43 (instead of 1) + 1.25 (instead of 1) + 1.11 (instead of 1) + 1.

So apart from the case where the divisor is a factor of n, we get an over-estimate for each quotient using the harmonic series.

So the over-estimate (in this case for n = 10) is obtained by summing the remainders for each quotient (obtained by dividing 10 by each number from 1 up to and including 10).

This sum = . 33 + .5 + .67 + .43 +.25 + .11 = 2.29

So the average remainder up to n = 10 = .229.

Now, as n gets larger it would be reasonable to assume that remainders would vary in a fairly predictable fashion over a range of values from 0 to 1.

Therefore the average remainder would be roughly .5.

Therefore the estimate of the average sum of divisors of n (as n increases without limit) would be overstated approximately by .5 using the sum for the harmonic series.

In fact, through using the two formulae, already obtained 1) for the sum of the harmonic series and 2) the average no. of divisors of n respectively, we can obtain a precise estimate for the average remainder.

Thus when we subtract 2) from 1) to get the averge value of the remainders,

= log n + γ – (log n – 1 + 2γ) = log n + γ – log n + 1 – 2γ = 1 – γ.

Therefore, with respect to a number n (when n can increase without limit), when it is divided successively by all numbers from 1 to n (inclusive), the average value with respect to the remainders of the quotients arising ~ 1 – γ.

Now the harmonic series relates to the value of the Riemann zeta function for s = 1.

Thus ζ(1) = 1 + 1/2 + 1/3 + 1/4 + .............

Remarkably, the result which we have obtained for the average remainder of these quotients can be related to all other values of the Riemann zeta function (from 2 to n).

So,

1 – γ = {ζ(2) – 1}/2 + {ζ(3) – 1}/3 + {ζ(4) – 1}/4 + {ζ(5) – 1}/5 +…