I'm trying to evaluate:∑n=0∞(−1)nln⁡2+2n1+2n

We have the series expansion of the digamma function: ψ(z)=−γ+∑n=0∞1n+1−1n+z Integrating both sides gives ln⁡Γ(z)=(1−z)γ+∑n=0∞z−1n+1+ln⁡[n+1n+z] Hence we have ln⁡Γ(1)+ln⁡Γ(14)−ln⁡Γ(12)−ln⁡Γ(34)=∑n=0∞ln⁡[(n+12)(n+34)(n+1)(n+14)] ln⁡[Γ(14)Γ(34)]−12ln⁡(π)=∑n=0∞ln⁡[(4n+2)(4n+3)(4n+4)(4n+1)]

S=∑n=0∞(−1)nlog⁡[2n+22n+1] =log⁡∏n=0∞[(4n+2)(4n+3)(4n+1)(4n+4)] =log⁡∏n=0∞[(1+24n)(1+34n}{(1+14n)(1+44n)} =log⁡∏n=0∞[(1+24n)(1+34n)(1+1n)14(1+1n)24(1+1n)34(1+14n)] =log⁡[Γ(14)Γ(24)Γ(34)]

We have the series expansion of the digamma function: ψ(z)=−γ+∑n=0∞1n+1−1n+z Integrating both sides gives ln⁡Γ(z)=(1−z)γ+∑n=0∞z−1n+1+ln⁡[n+1n+z] Hence we have ln⁡Γ(1)+ln⁡Γ(14)−ln⁡Γ(12)−ln⁡Γ(34)=∑n=0∞ln⁡[(n+12)(n+34)(n+1)(n+14)] ln⁡[Γ(14)Γ(34)]−12ln⁡(π)=∑n=0∞ln⁡[(4n+2)(4n+3)(4n+4)(4n+1)]

S=∑n=0∞(−1)nlog⁡[2n+22n+1] =log⁡∏n=0∞[(4n+2)(4n+3)(4n+1)(4n+4)] =log⁡∏n=0∞[(1+24n)(1+34n}{(1+14n)(1+44n)} =log⁡∏n=0∞[(1+24n)(1+34n)(1+1n)14(1+1n)24(1+1n)34(1+14n)] =log⁡[Γ(14)Γ(24)Γ(34)]

"I'm trying to evaluate:∑n=0∞(−1)nln⁡2+2n1+2n" was answered by our community

Jasper Carlson

Answered question

2022-02-22

I'm trying to evaluate:

\sum_{n = 0}^{\infty} {(- 1)}^{n} \ln \frac{2 + 2 n}{1 + 2 n}

Answer & Explanation

Pregazzix2a

Beginner2022-02-23Added 9 answers

We have the series expansion of the digamma function:

ψ (z) = - γ + \sum_{n = 0}^{\infty} \frac{1}{n + 1} - \frac{1}{n + z}

Integrating both sides gives

\ln Γ (z) = (1 - z) γ + \sum_{n = 0}^{\infty} \frac{z - 1}{n + 1} + \ln [\frac{n + 1}{n + z}]

Hence we have

\ln Γ (1) + \ln Γ (\frac{1}{4}) - \ln Γ (\frac{1}{2}) - \ln Γ (\frac{3}{4}) = \sum_{n = 0}^{\infty} \ln [\frac{(n + \frac{1}{2}) (n + \frac{3}{4})}{(n + 1) (n + \frac{1}{4})}]

\ln [\frac{Γ (\frac{1}{4})}{Γ (\frac{3}{4})}] - \frac{1}{2} \ln (π) = \sum_{n = 0}^{\infty} \ln [\frac{(4 n + 2) (4 n + 3)}{(4 n + 4) (4 n + 1)}]

Vanesa Burrows

Beginner2022-02-24Added 2 answers

S = \sum_{n = 0}^{\infty} {(- 1)}^{n} \log [\frac{2 n + 2}{2 n + 1}]

= {\log \prod}_{n = 0}^{\infty} [\frac{(4 n + 2) (4 n + 3)}{(4 n + 1) (4 n + 4)}]

= {\log \prod}_{n = 0}^{\infty} [\frac{(1 + \frac{\frac{2}{4}}{n}) (1 + \frac{\frac{3}{4}}{n}} {(1 + \frac{\frac{1}{4}}{n}) (1 + \frac{\frac{4}{4}}{n})}}{}

= {\log \prod}_{n = 0}^{\infty} [\frac{(1 + \frac{\frac{2}{4}}{n}) (1 + \frac{\frac{3}{4}}{n}) {(1 + \frac{1}{n})}^{\frac{1}{4}}}{{(1 + \frac{1}{n})}^{\frac{2}{4}} {(1 + \frac{1}{n})}^{\frac{3}{4}} (1 + \frac{\frac{1}{4}}{n})}]

= \log [\frac{Γ (\frac{1}{4})}{Γ (\frac{2}{4}) Γ (\frac{3}{4})}]

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