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

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]

enter image description here

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]

enter image description here

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]

enter image description here

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.

POSTED BY: Marvin Ray Burns
4 Replies

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 BY: Marvin Ray Burns

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 BY: Marvin Ray Burns

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 BY: Daniel Lichtblau
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} *)
POSTED BY: Greg Hurst
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