Infinite sum of logs puzzle Here is a neat infinite sum

Ernest Ryland

Ernest Ryland

Answered question

2022-01-22

Infinite sum of logs puzzle
Here is a neat infinite sum puzzle:
Prove the following is true when |x|<1
ln(1x)=ln(1+x)+ln(1+x2)+ln(1+x4)ln(1+x2k)
ln(1x)=k=0ln(1+x2k)

Answer & Explanation

xandir307dc

xandir307dc

Beginner2022-01-22Added 35 answers

ln(1x)=k=0ln(1+x2k)11x=k=0(1+x2k)
1=(1x)limnk=0n(1+x2k)
1=(1x2)limnk=0n(1+x2k)
1=limn(1x2n+1)
William Appel

William Appel

Beginner2022-01-23Added 44 answers

Exponentiate both sides to get 11x=(1+x)(1+x2) By uniqueness of the binary representation of a nonnegative integer we find that the xn coefficient on the right side is 1 for all n, so equality follows by power series representation of the left side.
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

Lets call the right hand side ln(A) Take the exponential of ln(A) and expand it with by 1x1x see the telescope product on the right side to get: A=11xlimk(1x2k+1)=11x Take the logarithm of A to show that it is equal to ln(1x).

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