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# What happens when trying to sum (I)^k (k^(1/k) - I)

Posted 10 years ago
 Since the MRB constant is Sum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}],  it got me wondering what the divergent Sum[(I)^n (n^(1/n) - I), {n, 1, Infinity}],  where I is the imaginary unit in 2 places above, does. I found out the sum has 4 accumulation points. Here is how I deduced that: l = N[Accumulate[Table[(I)^k (k^(1/k) - I), {k, 100}]]]; ListPlot[(Tooltip[{Re[#1], Im[#1]}] &) /@ l, AspectRatio -> 1]  In[24]:= Table[l[[-x]], {x, 1, 4}] Out[24]= {-0.165382 - 0.258948 I, -1.21251 + 0.741052 I, -0.21251 + 1.78856 I, 0.835387 + 0.788561 I} In[25]:= l = N[Accumulate[Table[(I)^k (k^(1/k) - I), {k, 1000}]]]; In[26]:= ListPlot[(Tooltip[{Re[#1], Im[#1]}] &) /@ l, AspectRatio -> 1]  In[27]:= Table[l[[-x]], {x, 1, 4}] Out[27]= {-0.185294 - 0.238851 I, -1.19223 + 0.761149 I, -0.192226 + 1.76809 I, 0.814718 + 0.768086 I} In[28]:= l = N[Accumulate[Table[(I)^k (k^(1/k) - I), {k, 10000}]]]; In[29]:= ListPlot[(Tooltip[{Re[#1], Im[#1]}] &) /@ l, AspectRatio -> 1]  In[30]:= Table[l[[-x]], {x, 1, 4}] Out[30]= {-0.188296 - 0.235846 I, -1.18922 + 0.764154 I, -0.189218 + 1.76508 I, 0.811704 + 0.765075 I} In[39]:= -0.18829645859705124 - 0.23584620227704534 I - (-1.189217916916732 + 0.7641537977229609 I) Out[39]= 1.00092 - 1. I In[40]:= -0.18921791691673207 + 1.7650753382290296 I - (0.8117037057913876 + 0.7650753382290443 I) Out[40]= -1.00092 + 1. I In[41]:= -1.189217916916732 + 0.7641537977229609 I - (-0.18921791691673207 + 1.7650753382290296 I) Out[41]= -1. - 1.00092 I,  where the 1.00092 either eventually goes to 1 as n goes to infinity or it could be a separate (probably new) constant that the accumulation points differ by.
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Posted 10 years ago
 I tested Sum[(I)^n (n^(1/n) ), {n, 1, Infinity}] and Sum[(I)^n (n^(1/n) +I), {n, 1, Infinity}] and they both also seem to have 4 accumulation points that differ by 1 + 1I.
Posted 10 years ago
 Here is a brute force method that shows me Chip Hust's analysis is spot on. Table[NSum[(I)^n (n^(1/n) - I), {n, 1, 4^12 + x}, WorkingPrecision -> 30] - NSum[(I)^n (n^(1/n) - I), {n, 1, 4^12 + x + 1}, WorkingPrecision -> 30], {x, 4}]1 I (*{1.000000991555 - 1.000000000000 I, 1.000000000000 + 1.000000991555 I, -1.000000991555 + 1.000000000000 I, -1.000000000000 - 1.000000991555 I}*) Compared with my original post, the non 1 sum is shown going to 1 as n goes to infinity, where 1.00092>1..000000991555>1.
Posted 10 years ago
 Q: What happens to partial sums in the standard MRB definition, if that -1 is removed from (n^(1/n) - 1).A: Even partial sums still go to MRB. Odd ones are short a one, hence go to MRB-1.I think a similar analysis will give some insight into the four-headedness you observe on replacing -1 with i or -I.
Posted 10 years ago
 Expanding on that idea, we can take real and imaginary parts: re = Refine[ComplexExpand[Re[I^n (n^(1/n) - I)]], n > 0] (* n^(1/n) Cos[(n ?)/2] + Sin[(n ?)/2] *) im = Refine[ComplexExpand[Im[I^n (n^(1/n) - I)]], n > 0] (* n^(1/n) Sin[(n ?)/2] - Cos[(n ?)/2] *) These trig functions have period 4, hence your 4 accumulation points. We can simplify your summand mod 4 to see all four MRB-like constant summands: held = re + I im /. {n^(1/n) :> HoldForm[n^(1/n)]}; mod4cases = FullSimplify[held /. {n -> 4k + #}, k ? Integers && k > 0]& /@ Range[0, 3] (* {(4k)^(1/(4k)) - I, I (4k+1)^(1/(4k+1)) + 1, -(4k+2)^(1/(4k+2)) + I, -I (4k+3)^(1/(4k+3)) - 1} *)