Convergence of Some Divergent Series!

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1 Convergence of Some Divergent Series! T. Muthukumar 9 Jun 04 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, n(n + k = n = is finite and k = = +. 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, k and ( k are diverging, while k s s C and R(s >, is converging. The convergence of a series is defined by the convergence of its partial sum. For instance, k

2 diverges because its partial sum s n := k = n(n + diverges, as n. The geometric series z k = z, ( for all z <, because its partial sum s n := zn z n z. Note that the map T (z = is well-defined between T : C \ {} C. z In fact, T is a holomorphic (complex differentiable functions and, hence, analytic. The analytic function T exists for all complex numbers except z = and, inside the unit disk {z C z < }, T is same as the power series in (. Thus, T may be seen as an analytic continuation of outside the unit disk, except at z =. 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 z k. In this sense, setting z = in T (z, the divergent series may be seen as taking the finite value k = =. Similarly, setting z = in T (z, the oscillating divergent series may be seen as taking the finite value ( k = ( =. z k

3 This sum coincides with the Cesàro sum which is a special kind of convergence for series which do not diverge to ±. For instance, k diverges in Cesàro sum too. The Cesàro sum of a series is defined as the Observe that the sequence n lim n n i= ( i a k ( i a k. i= ( k. Since T is not defined on, we have not assigned any finite value to the divergent series The Riemann zeta function ζ(z := converges for z C with R(z >. Note that for z C, = e z ln(k, where ln is the real logarithm. If z = σ + it then = k σ e it ln(k and = k σ since e it ln(n =. Therefore, the infinite series converges for all z C such that R(z > (i.e. σ > because it is known that k σ converges for all σ > and diverges for all σ. The Riemann zeta function is a special case of the Dirichlet series D(z := a k 3

4 with a k = for all k. Note that for z =, we get the Basel series ζ( := k = π 6 and ζ(3 = 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 (Euler. For all k N, k+ (πk ζ(k = ( (k! B k where B k is the k-th Bernoulli number. If ζ(z is well-defined for R(z > then z ζ(z is also well-defined for R(z >. Therefore, ζ(z( z = ( k n z =: η(z. Observe that η(z is also a Dirichlet series. It can be shown that ( z η(z is analytic for all R(z > 0 and R(z. Thus, we have analytically continued ζ(z for all R(z > 0 and R(z. There is a little work to be done on the zeroes of z but is fixable. In the strip 0 < R(z <, called the critical strip, the Riemann zeta function satisfies the relation ( zπ ζ(z = z (π z sin Γ( zζ( z. This relation is used to extend the Riemann zeta function to z with nonpositive real part, thus, extending to all complex number z. Setting z = in the relation yields ζ( = π sin ( π Since ζ( is extension of the series Γ(ζ( = π ( π 6 =. 4

5 one may think of ζ( as the finite value corresponding to The value of ζ(0 is obtained limiting process because it involves ζ( which is not defined. However, there is an equivalent formulation of Riemann zeta function for all z C \ {} as ζ(z = z k+ k ( k ( i (i + z i with a simple pole and residue at z =. Recall that the analytic continuation T of the geometric series is not defined for z = and, hence, we could not assign a finite value to But, setting z = 0 above, T ( = ζ(0 = i=0 T ( = k+ k ( k ( i δ 0k = i = k+ i=0 where δ 0k is the Kronecker delta defined as { k = 0 δ 0k = 0 k 0. The harmonic series n= corresponds to ζ( which is not defined. The closest one can conclude about ζ( is that ( lim n k ln(n = γ n where γ is the Euler-Mascheroni constant which has the value γ =

ζ (s) = s 1 s {u} [u] ζ (s) = s 0 u 1+sdu, {u} Note how the integral runs from 0 and not 1.

ζ (s) = s 1 s {u} [u] ζ (s) = s 0 u 1+sdu, {u} Note how the integral runs from 0 and not 1. Problem Sheet 3. From Theorem 3. we have ζ (s) = + s s {u} u+sdu, (45) valid for Res > 0. i) Deduce that for Res >. [u] ζ (s) = s u +sdu ote the integral contains [u] in place of {u}. ii) Deduce that for