Determine if the following series or improper integrals are convergent or divergent:
Solution
Here is a complex-analytic proof.
For zD=C{0,1}, let
R(z)=1log2z
where the sum is taken over all branches of the logarithm. Each point in D has a neighbourhood
on which the branches of log(z) are analytic. Since the series converges uniformly away from
z=1, R(z)is analytic on D.
Now a few observations:
(i) Each term of the series tends to 0 as z0. Thanks to the uniform convergence this implies that
the singularity at z=0 is removable and we can set R(0)=0.
(ii) The only singularity of R is a double pole at z=1 due to the contribution of the principal
branch of logz. Moreover, limz1(z1)2R(z)=1.
(iii) R(1/z)=R(z).
By (i) and (iii) R is meromorphic on the extended complex plane, therefore it is rational. By (ii)
the denominator of R(z) is (z1)2. Since R(0)=R()=0, the numerator has the form az. Then (ii)
implies a=1, so thatR(z)=z(z1)2.
Now, setting z=e2iw yieldsn=1(wn)2=2sin2(w)
down vote
Here is a complex-analytic proof.
For zD=C{0,1}, let
R(z)=1log2z
where the sum is taken over all branches of the logarithm. Each point in D has a
neighbourhood on which the branches of log(z) are analytic. Since the series converges
uniformly away from z=1, R(z)is analytic on D.
Now a few observations:
(i) Each term of the series tends to 0 as z0. Thanks to the uniform convergence this implies
that the singularity at z=0 is removable and we can set R(0)=0.
(ii) The only singularity of R is a double pole at z=1 due to the contribution of the principal
branch of logz. Moreover, limz1(z1)2R(z)=1.
(iii) R(1/z)=R(z).
By (i) and (iii) R is meromorphic on the extended complex plane, therefore it is rational. By (ii)
the denominator of R(z) is (z1)2. Since R(0)=R()=0, the numerator has the form az. Then (ii)
implies a=1, so thatR(z)=z(z1)2.
Now, setting z=e2iw yieldsn=1(wn)2=2sin2(w)
which implies thatk=01(2k+1)2=28,
and the identity (2)=2/6 follows..
Determine if the following series or improper integrals are converge.pdf
1. Determine if the following series or improper integrals are convergent or divergent:
Solution
Here is a complex-analytic proof.
For zD=C{0,1}, let
R(z)=1log2z
where the sum is taken over all branches of the logarithm. Each point in D has a neighbourhood
on which the branches of log(z) are analytic. Since the series converges uniformly away from
z=1, R(z)is analytic on D.
Now a few observations:
(i) Each term of the series tends to 0 as z0. Thanks to the uniform convergence this implies that
the singularity at z=0 is removable and we can set R(0)=0.
(ii) The only singularity of R is a double pole at z=1 due to the contribution of the principal
branch of logz. Moreover, limz1(z1)2R(z)=1.
(iii) R(1/z)=R(z).
By (i) and (iii) R is meromorphic on the extended complex plane, therefore it is rational. By (ii)
the denominator of R(z) is (z1)2. Since R(0)=R()=0, the numerator has the form az. Then (ii)
implies a=1, so thatR(z)=z(z1)2.
Now, setting z=e2iw yieldsn=1(wn)2=2sin2(w)
down vote
Here is a complex-analytic proof.
For zD=C{0,1}, let
R(z)=1log2z
where the sum is taken over all branches of the logarithm. Each point in D has a
neighbourhood on which the branches of log(z) are analytic. Since the series converges
uniformly away from z=1, R(z)is analytic on D.
Now a few observations:
(i) Each term of the series tends to 0 as z0. Thanks to the uniform convergence this implies
that the singularity at z=0 is removable and we can set R(0)=0.
(ii) The only singularity of R is a double pole at z=1 due to the contribution of the principal
branch of logz. Moreover, limz1(z1)2R(z)=1.
(iii) R(1/z)=R(z).
2. By (i) and (iii) R is meromorphic on the extended complex plane, therefore it is rational. By (ii)
the denominator of R(z) is (z1)2. Since R(0)=R()=0, the numerator has the form az. Then (ii)
implies a=1, so thatR(z)=z(z1)2.
Now, setting z=e2iw yieldsn=1(wn)2=2sin2(w)
which implies thatk=01(2k+1)2=28,
and the identity (2)=2/6 follows.
