Get the answer to "Prove that:limm→∞∑k=1m2−kk=log⁡2"

cindakayumn

Answered question

2022-02-24

Prove that:

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

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

\sum_{k = 1}^{\infty} \frac{2^{- k}}{k}

= \sum_{k = 1}^{\infty} 2^{- k} \int_{0}^{1} x^{k - 1} d x

= \int_{0}^{1} \sum_{k = 1}^{\infty} 2^{- k} x^{k - 1} d x

= \frac{1}{2} \int_{0}^{1} \sum_{k = 1}^{\infty} {(\frac{x}{2})}^{k - 1} d x

= \frac{1}{2} \int_{0}^{1} \frac{1}{1 - \frac{x}{2}} d x

= {[- \ln | 1 - \frac{x}{2} |]}_{0}^{1}

= \ln 2

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