Simplify log expression with infinite series \log x-\sum^{\infty}_{i=1}\frac{x^i}{i}

Deven Livingston

Deven Livingston

Answered question

2022-04-25

Simplify log expression with infinite series logx{i=1}xii

Answer & Explanation

narratz5dz

narratz5dz

Beginner2022-04-26Added 13 answers

Hint: taylor series:
log(1x)=i=1xii=xx22x33 for |x|<1

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