Well, It is definitely not gonna work when you write

ln

instead of

Log

!

Here is the correct code to solve this problem:

ClearAll[GetSum]
GetSum[w_?NumberQ,c_?NumberQ]:=NSum[1/2^n*(Log[w+2^n-c]-Log[w]),{n,1,Infinity}]

w=1000000;
Plot[GetSum[w,c],{c,1,30}]

c/.FindRoot[GetSum[w,c],{c,20,15,25}]

Best I can tell, Sum has no real chance to handle that input, and so Solve and even NSolve are likely to fail. I'd suggest making a function that can be evaluated numerically by NSum and using FindRoot on the result. Something like below.

Define w and the NSum.

w = 1000000;
util[c_?NumberQ] :=  NSum[1/2^n*(Log[w + 2^n - c] - Log[w]), {n, 1, Infinity}]

Now find the value of c that zeroes the sum.

FindRoot[util[c] == 0, {c, 2}]

(* Out[1148]= {c -> 20.8753540254} *)

This is close to the reference value of 20.84. Playing with various options brings it a hair closer, to 20.874. Maybe I'm not using the best options though. Or maybe the reference value had a typo, hard to say.

Hi Daniel,

Thanks for your reply.

I'm sorry, I mistakenly wrote 100 in my initial wealth rather than 1,000,000 which is what I tried in Mathematica and which didn't give a numerical result.

I changed the value to the correct 1,000,000 five minutes after my initial post, but I guess you only saw the first incorrect version. Sorry about that.

Here is the mathematica code that I tried to run with no numerical result:

w = 1000000;

Solve[Sum[1/2^n*(ln[w + 2^n - c] - ln[w]), {n, 1, Infinity}] == 0, c]

Best regards, Keith