Expert answers to "How to show that that ∑n=1∞(13n−1+13n−2−23n)=ln⁡(3)"

Jaeden Frederick

Answered question

2022-01-23

How to show that that

\sum_{n = 1}^{\infty} (\frac{1}{3 n - 1} + \frac{1}{3 n - 2} - \frac{2}{3 n}) = \ln (3)

Jude Carpenter

Beginner2022-01-24Added 9 answers

Use the fact that

\frac{1}{m + 1} = \int_{0}^{1} x^{m} d x

to get

\sum_{n = 1}^{\infty} \int_{0}^{1} (x^{3 n - 2}} + x^{3 n - 3} - 2 x^{3 n - 1}) d x

= \int_{0}^{1} \sum_{n = 1}^{\infty} (x^{3 n - 2} + x^{3 n - 3} - 2 x^{3 n - 1}) d x

= \int_{0}^{1} \frac{x + 1 - 2 x^{2}}{1 - x^{3}} d x = \int_{0}^{1} \frac{2 x + 1}{x^{2} + x + 1} d x = \log (x^{2} + x + 1) ∣_{0}^{1} = \log 3

kumewekwah0

Beginner2022-01-25Added 14 answers

Let

S_{N} = \sum_{n = 1}^{N} \frac{1}{n}

be the n-th partial sum of the harmonic series. Then

\sum_{n = 1}^{N} (\frac{1}{3 n - 1} + \frac{1}{3 n - 2} - \frac{2}{3 n}) = \sum_{n = 1}^{N} (\frac{1}{3 n - 1} + \frac{1}{3 n - 2} + \frac{1}{3 n}) - \sum_{n = 1}^{N} \frac{1}{n}

= - s_{3 N} - s_{N}

It is well-known that

s_{N} = \ln N + γ + O (\frac{1}{N})

, hence

\sum_{n = 1}^{N} (\frac{1}{3 n - 1} + \frac{1}{3 n - 2} - \frac{2}{3 n})

= \ln (3 N) - \ln N + O (\frac{1}{N})

= \ln 3 + O (\frac{1}{N})

RizerMix

Expert2022-01-27Added 656 answers

Lets

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