MATLAB/Divergent series investigations

Shown below are the 15 terms of the series for &epsilon; equal to 0, .001, and .01, respectively. The series converges to -1/12 in the limit that &epsilon;>0 vanishes. S(&epsilon;=.000) = 1.0 + 2.0 + 3.0 + 4.0 + 5.0 + 6.0 + 7.0 + 8.0 + 9.0 + 10.0 + 11.0 + 12.0 + 13.0 + 14.0 + 15.0 + ... S(&epsilon;=.001) = 1.0 + 2.0 + 3.0 + 4.0 + 5.0 + 6.0 + 7.0 + 7.9 + 8.9 + 09.9 + 10.9 + 11.9 + 12.8 + 13.8 + 14.8 + ... S(&epsilon;=.010) = 1.0 + 2.0 + 2.9 + 3.8 + 4.8 + 5.6 + 6.5 + 7.4 + 8.2 + 09.0 + 09.8 + 10.6 + 11.3 + 12.1 + 12.8 + ...

This code runs on both MATLAB and Octave. It sums,
 * $$S(N,\epsilon)=\sum_{n=1}^N n z^n$$ where, $$z=e^{-\epsilon+i\epsilon} \approx 1 \ \mathrm{if} \ \epsilon<<1$$.

The numerical simulation suggests that for real $$\epsilon$$:


 * $$\lim_{\epsilon\to 0} \sum_{n=1}^\infty n e^{-\epsilon n}\cos(\epsilon n)=-\frac {1}{12}$$

Warning: This result does not hold if $$\epsilon$$ is a complex number that goes to zero, unless the real and imaginary parts are constrained to be equal in magnitude. This would be a significant simulation if the series converges to -1/12 for all ratios of real to imaginary parts of &epsilon;. Nevertheless, it is curious that a real to imaginary ratio of one yields the well known value, 1/12&asymp;.083333

Questions

 * 1) Verify that $$\int_0^\infty x e^{-\epsilon x}\cos(\epsilon x) dx = 0$$ if &epsilon; > 0
 * 2) Why is this interesting?
 * 3) Can Octave do symbolic integration?

Output of MATLAB code
The code that follows gave this output:

If z= 0.990000332+0.009900333i the sum from 1 to 4000 of n*z^n is -8.333333332e-02 + i 4.999999999e+03

MATLAB/Octave code
% This code is designed to investigate the "astonishing" claim that %  % 1 + 2 + 3 + 4 + 5 + ... = -1/12  %  % It numerically calculates % %   1*z + 2*z^2 + 3*z^3 + 4*z^4 + 5*z^5 + ... N*z^N %  % where N is large and z is nearly equal to, 1.0 + 0.0i clear;clc; close all;% clear memory and command window N=4000; % sets the number of terms in the expansion shortN=20; % how far we plot to show initial behaviour of the series alpha=.01; %the real part of z beta=.01; epsilon = -alpha+1*i*beta; % The result is astonishing if alpha=beta z=exp(epsilon); theta=atan(abs(imag(z)/real(z)));%is the argument of z period=round(2*pi/theta); %We estimate the limit as N->infty by averaging over one period n=1:N; % the partial sums shall be plotted against these integers seq=linspace(0,0,N); % preallocate elements of the sequence {1,2,3...,N} parsum=linspace(0,0,N);% preallocate the partial sums: {1, 1+2, 1+2+3, ...} seq(1)=1*z; % first element of sequence parsum(1)=seq(1); % first partial sum %All the partial sums are stored in parsum, all the terms in seq for count=2:N seq(count)=count*z^count; parsum(count)=parsum(count-1)+seq(count); end parsumR=real(parsum); % we plot only the real part subplot(3,1,1) plot(n,parsumR,'.k',[0, N],[-1/12,-1/12],'MarkerSize',6.0,'Linewidth',1.5); axis([0,shortN,0,shortN*(shortN+1)/2]); %title('(a) The initial partial sums proceed almost as 1+2+3+4+...'); % The titles do not print properly on Octave and had to be commented out %plots the first terms (should go as n*(n+1)/2) Verify with parabola: hold on; plot(n,n.*(n+1)/2,'-r',[0, N],[-1/12,-1/12]); hold off; subplot(3,1,2) plot(n,parsumR,'.k',[0, N],[-1/12,-1/12],'MarkerSize',8.0); %title('(b) The small deviation from z=1 introduces decay and oscillation'); %plots all N terms subplot(3,1,3) plot(n,parsumR,'.k',[0, N],[-1/12,-1/12],'MarkerSize',8.0); axis([0,N,-.25,.25]); %title('(c) The last terms show convergence to approximately -1/12'); %plots the partial sums for the largest values of n. final=parsum(N); fprintf('If z= \n%11.9f+%11.9fi ',real(z),imag(z) ); fprintf('the sum from 1 to %d of n*z^n is \n',N ); fprintf(' %11.9e + i %11.9e\n',real(final),imag(final));%this works %%%%Save the real part of the sequence for count=1:N Rseq(count)=real(seq(count)); end