Average of Sigma(n)/n
One of the simplest proofs of the fact that the average "order" of sigma(n)/n
is equal to zeta(2) = pi^2/6 (i.e., the sum of the inverse squares) is by
counting the lattice points in certain plane regions (c.f., Hardy and Wright),
but there are some other interesting approaches as well. 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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