Prove that: \lim_{m\to\infty}\sum_{k=1}^m\frac{2^{-k}}{k}=\log 2

cindakayumn

cindakayumn

Answered question

2022-02-24

Prove that:
limmk=1m2kk=log2

Answer & Explanation

Zachary Burton

Zachary Burton

Beginner2022-02-25Added 1 answers

By comparison with the geometric series with 1/2 as common ratio, the series in the question converges absolutely, so we can write
k=12kk
=k=12k01xk1dx
=01k=12kxk1dx
=1201k=1(x2)k1dx
=120111x2dx
=[ln|1x2|]01
=ln2.

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