Evaluate ∑ ℓ ≥ 1 1 ℓ ( 2 ℓ − 1 )

I am answering the question in the comments. It has to be shown that − ∑ k = 1 ∞ ln ⁡ ( 1 − 2 − k ) &lt; ln ⁡ 4.Note that for 0 &lt; x &lt; 1 2 − ln ⁡ ( 1 − x ) = x + x 2 2 + x 3 ∫ 0 1 t 2 1 − x t d t ≤ x + x 2 2 + x 3 ∫ 0 1 t 2 1 − t / 2 d t &lt; x + x 2 2 + 2 x 3 3 .Then − ∑ k = 1 ∞ ln ⁡ ( 1 − 2 − k ) &lt; ∑ k = 1 ∞ 2 − k + 1 2 ∑ k = 1 ∞ 4 − k + 2 3 ∑ k = 1 ∞ 8 − k = 53 42 &lt; 1.3 &lt; ln ⁡ 4.

Expert answers to "Evaluate ∑ ℓ ≥ 1 1 ℓ (..."

College
Calculus and Analysis
Integral Calculus
Integrals

Cooper Doyle

Answered question

2022-07-07

Evaluate

\sum_{ℓ \geq 1} \frac{1}{ℓ (2^{ℓ} - 1)}

Answer & Explanation

bap1287dg

Beginner2022-07-08Added 13 answers

I am answering the question in the comments. It has to be shown that

- \sum_{k = 1}^{\infty} \ln (1 - 2^{- k}) < \ln 4.

Note that for

0 < x < \frac{1}{2}

- \ln (1 - x) = x + \frac{x^{2}}{2} + x^{3} \int_{0}^{1} \frac{t^{2}}{1 - x t} d t \leq x + \frac{x^{2}}{2} + x^{3} \int_{0}^{1} \frac{t^{2}}{1 - t / 2} d t < x + \frac{x^{2}}{2} + \frac{2 x^{3}}{3} .

Then

- \sum_{k = 1}^{\infty} \ln (1 - 2^{- k}) < \sum_{k = 1}^{\infty} 2^{- k} + \frac{1}{2} \sum_{k = 1}^{\infty} 4^{- k} + \frac{2}{3} \sum_{k = 1}^{\infty} 8^{- k} = \frac{53}{42} < 1.3 < \ln 4.

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