Thursday, 19 June 2014

`Convergence' of Some Divergent Series

‘Convergence’ of Some Divergent Series!

The topic of this article, the idea of attaching a finite value to divergent series, is no longer a purely mathematical exercise. These finite values of divergent series have found application in String theory and quantum field theory (Casimir effect).

The finite sum of real/complex numbers is always finite. The infinite sum of real/complex numbers can be either finite or infinite. For instance,

k=1nk = 1 + 2 + + n = n(n + 1) 2

is finite and

k=1k = 1 + 2 + 3 + = +.

If an infinite sum has finite value it is said to converge, otherwise it is said to diverge. Divergence do not always mean it grows to ±. For instance,

k=1k, k=11 k and  k=0(1)k

are diverging, while

k=1 1 kss  and (s) > 1,

is converging. The convergence of a series is defined by the convergence of its partial sum. For instance,

k=1k

diverges because its partial sum

sn := k=1nk = n(n + 1) 2

diverges, as n . The geometric series

k=0zk = 1 1 z, (1)

for all |z| < 1, because its partial sum

sn := zn 1 z 1 n 1 z 1.

Note that the map T(z) = 1 1z is well-defined between T : {1} . In fact, T is a holomorphic (complex differentiable) functions and, hence, analytic. The analytic function T exists for all complex numbers except z = 1 and, inside the unit disk {z |z| < 1}, T is same as the power series in (1). Thus, T may be seen as an analytic continuation of

k=0zk

outside the unit disk, except at z = 1. A famous unique continuation result from complex analysis says that the analytic continuation is unique. Thus, one may consider the value of T(z), outside the unit disk, as an ‘extension’ of the divergent series

k=0zk.

In this sense, setting z = 2 in T(z), the divergent series may be seen as taking the finite value

k=02k = 1 + 2 + 4 + 8 + 16 + = 1.

Similarly, setting z = 1 in T(z), the oscillating divergent series may be seen as taking the finite value

k=0(1)k = 1 1 (1) = 1 2.

This sum coincides with the Cesàro sum which is a special kind of convergence for series which do not diverge to ±. For instance,

k=1k

diverges in Cesàro sum too. The Cesàro sum of a series

k=1a k

is defined as the

lim n1 n i=1n k=1ia k .

Observe that the sequence

1 n i=1n k=0i1(1)k 1 2.

Since T is not defined on 1, we have not assigned any finite value to the divergent series

1 + 1 + 1 + .

The Riemann zeta function

ζ(z) := k=1 1 kz

converges for z with (z) > 1. Note that for z , kz = ez ln(k), where ln is the real logarithm. If z = σ + it then kz = kσeit ln(k) and |kz| = kσ since |eit ln(n)| = 1. Therefore, the infinite series converges for all z such that (z) > 1 (i.e. σ > 1) because it is known that k=1kσ converges for all σ > 1 and diverges for all σ 1.

The Riemann zeta function is a special case of the Dirichlet series

D(z) := k=1ak kz

with ak = 1 for all k. Note that for z = 2, we get the Basel series

ζ(2) := k=1 1 k2 = π2 6

and ζ(3) = 1.202056903159594 called the Apéry’s constant. In fact, the following result of Euler gives the value of Riemann zeta function for all even positive integers.

Theorem 1 (Euler). For all k ,

ζ(2k) = (1)k+1(2π)2k 2(2k)!B2k

where B2k is the 2k-th Bernoulli number.

If ζ(z) is well-defined for (z) > 1 then 21zζ(z) is also well-defined for (z) > 1. Therefore,

ζ(z)(1 21z) = k=1(1)k1nz =: η(z).

Observe that η(z) is also a Dirichlet series. It can be shown that (1 21z)η(z) is analytic for all (z) > 0 and (z)1. Thus, we have analytically continued ζ(z) for all (z) > 0 and (z)1. There is a little work to be done on the zeroes of 1 21z but is fixable. In the strip 0 < (z) < 1, called the critical strip, the Riemann zeta function satisfies the relation

ζ(z) = 2z(π)z1 sin zπ 2 Γ(1 z)ζ(1 z).

This relation is used to extend the Riemann zeta function to z with non-positive real part, thus, extending to all complex number z1. Setting z = 1 in the relation yields

ζ(1) = 1 2π2 sin π 2 Γ(2)ζ(2) = 1 2π2(1)π2 6 = 1 12 .

Since ζ(1) is extension of the series

k=1 1 kz

one may think of ζ(1) as the finite value corresponding to 1 + 2 + 3 + . The value of ζ(0) is obtained limiting process because it involves ζ(1) which is not defined. However, there is an equivalent formulation of Riemann zeta function for all z {1} as

ζ(z) = 1 1 21z k=0 1 2k+1 i=0k(1)ik i (i + 1)z

with a simple pole and residue 1 at z = 1.

Recall that the analytic continuation T of the geometric series is not defined for z = 1 and, hence, we could not assign a finite value to

T(1) = 1 + 1 + 1 + .

But, setting z = 0 above,

T(1) = ζ(0) = k=0 1 2k+1 i=0k(1)ik i = k=0 δ0k 2k+1 = 1 2

where δ0k is the Kronecker delta defined as

δ0k = 1k = 00 k 0.

The harmonic series

n=11 n

corresponds to ζ(1) which is not defined. The closest one can conclude about ζ(1) is that

lim n k=1n1 k ln(n) = γ

where γ is the Euler-Mascheroni constant which has the value γ = 0.57721566.