Average of Sigma(n)/n
Hardy and Wright's "Introduction to the Theory of Numbers" gives
a simple demonstration that the average "order" of sigma(n)/n is
equal to zeta(2) (the sum of the inverse squares) based a count
of the lattice points in a plane region, but there are some other
interesting approaches to this problem. For example, by the Taylor
series expansion of the natural log function, we have
/ x x^2 x^3 x^4 \
ln(1-x) = -( --- + --- + --- + --- + ... ) (1)
\ 1 2 3 4 /
for all |x| less than 1. Now, if we consider the sum of the
natural logs of 1-x, 1-x^2, 1-x^3, and so on, we have
oo x x^2 x^3 x^4 x^5 x^6
-SUM ln(1 - x^n) = --- + --- + --- + --- + --- + --- + ...
n=1 1 2 3 4 5 6
x^2 x^4 x^6 x^8 x^10 x^12
+ --- + --- + --- + --- + ---- + ---- + ...
1 2 3 4 5 6
x^3 x^6 x^9 x^12 x^15 x^18
+ --- + --- + --- + ---- + ---- + ---- + ...
1 2 3 4 5 6
x^4 x^8 x^12 x^16 x^20 x^24
+ --- + --- + ---- + ---- + ---- + ---- + ...
1 2 3 4 5 6
+ etc...
Combining terms by powers of x gives the nice identity
oo oo sigma(n)
- SUM ln(1 - x^n) = SUM -------- x^n (2)
n=1 n=1 n
If we divide both sides of this relation by the geometric series, i.e.,
the sum of x^n for n=1 to oo, we have
oo sigma(n)
SUM -------- x^n
1-x oo n=1 n
- --- SUM ln(1 - x^n) = ------------------ (3)
x n=1 oo
SUM x^n
n=1
Notice that the right side of this equation a geometrically weighted
average of all the values of sigma(n)/n from n=1 to infinity. Also,
as x approaches 1, this approaches the EVENLY weighted average out to
arbitrarily large n. This suggests that it would be interesting to
evaluate the limit of the left side as x goes to 1. This turns out
to be sort of a delicate operation, because if the summation on the
left side of (3) is evaluated for n=1 to N for any fixed N, the overall
quantity on on the left side goes to zero as x goes to 1. However, for
any fixed x less than 1, the quantity converges on a non-zero value as
N goes to infinity, and these limiting values converge on a certain
value as x goes to 1.
Anyway, by making x sufficiently close to 1 we can make the ratios of
successive terms in the left-hand summation arbitrarily close to 1,
so the summation can be approached by the integral in the limit as
x goes to 1. In other words
oo
oo /
lim SUM ln(1 - x^n) = | ln(1 - x^n) dn (4)
x->1 n=1 /
n=1
To evaluate the integral, let's make the substitution q = x^n, where
q goes from x to 0 as n goes from 1 to infinity. Noting that q can be
written as exp(nln(x)) we have
dq
-- = exp(n ln(x)) ln(x) = ln(x) x^n (5)
dn
which gives
1 1
dn = --------- dq = ------- dq
ln(x) x^n q ln(x)
With these substitutions the integral becomes
oo 0
/ 1 / ln(1-q)
| ln(1 - x^n) dn = ----- | ------- dq (6)
/ ln(x) / q
n=1 q=x
Recalling the power series expansion of ln(1-q), we have the nice
indefinite integral
/ ln(1-q) / q q^2 q^3 q^4 \
| ------- dq = - ( --- + --- + --- + --- + ... ) (7)
/ q \1^2 2^2 3^2 4^2 /
Evaluating this from q=x to q=0 gives
0
/ ln(1-q) x x^2 x^3 x^4
| ------- dq = --- + --- + --- + --- + ...
/ q 1^2 2^2 3^2 4^2
q=x
Substituting this back into equation (6), and from there back into
equation (3), we arrive at the expression for the evenly weighted
average of sigma(n)/n for all integers n:
1 - x / x x^2 x^3 x^4 \
lim - ------- ( --- + --- + --- + --- + ... )
x->1 x ln(x) \ 1^2 2^2 3^2 4^2 /
Since the ratio of -(1-x) to ln(x) goes to 1 as x approaches 1,
and the numerators inside the parentheses also go to 1, we have the
result
oo sigma(n)
SUM -------- x^n
n=1 n oo 1
lim ------------------ = SUM ---
x->1 oo n=1 n^2
SUM x^n
n=1
which shows, as expected, that the average value of sigma(n)/n for all
integers is zeta(2) = pi^2 / 6. This approach is certainly much less
economical than the simple lattice-point derivation presented in Hardy
and Wright, but it does provide the opportunity to relate some common
power series expansions to arithmetic functions. For example, from the
expansions of (1 - 1/x)ln(1-x^n) we see that the ratio sigma(n)/n equals
the sum of the numbers in the first n columns of the following array:
1 -1/2 -1/6 -1/12 -1/20 -1/30 -1/42 -1/56 -1/72 -1/90 -1/110
1 -1 1/2 -1/2 1/3 -1/3 1/4 -1/4 1/5 -1/5
1 -1 1/2 -1/2 1/3 -1/3
1 -1 1/2 -1/2
1 -1 1/2 -1/2
1 -1
1 -1
1 -1
1 -1
1 -1
1
This is related to the interesting fact that the series expansion of
the left side of (3) is
1 1 5 11 24
1 + - x - - x^2 + -- x^3 - -- x^4 + -- x^5 - ...
2 6 12 20 30
where the coefficient of x^k is sigma(k)/k - sigma(k-1)/(k-1). This
converges for all x less than 1, whereas for x=1 the partial sums
are just the successive values of sigma(k)/k, so it never converges.
Nevertheless, the limit of the convergent values as x approaches 1
is (pi^2)/6.
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