Order of Summation for the Alternating Harmonic Series

Helmut  Knaust - Department of Mathematical Sciences, University of Texas at El Paso, hknaust@utep.edu
9/15/2014

We will be summing the alternating harmonic series in its normal order  (“red”):

1-(1/2)+(1/3)-(1/4)+(1/5)....

and then in a rearranged order ("blue"):

1+(1/3)-(1/2)  +  (1/5)+(1/7)-(1/4)   +   (1/9)+(1/11)-(1/6)   +   (1/13)+(1/15)-(1/8)   + ...

Both series converge, albeit to different limits.

Clear[a]

red[m_]:=Sum[a[2k-1]-a[2k],{k,1,3m}]
red[1]
red[2]

blue[m_]:=Sum[a[4k-3]+a[4k-1]-a[2k],{k,1,2m}]
blue[1]
blue[2]

a[k_]=1/k;

The plot below lists the partial sums of the “red” and “blue” series:

z=50;
pl1=ListPlot[Table[{6m,red[m]},{m,1,z,Ceiling[z/100]}],PlotStyle->{Thick,Red}];
pl2=ListPlot[Table[{6m,blue[m]},{m,1,z,Ceiling[z/100]}],PlotStyle->{Thick,Blue}];
Show[{pl1,pl2},PlotRange->All]

Summing in the “blue” order yields one and a half times the result of summing in the standard “red” order:

blue[100]/red[100]//N

Another Example

For a[k]=(1/(Sqrt[k])):