Expert Answer to "Compute the limit:∑k=1∞ln⁡(1+14k2)"

Sarah-Louise Prince

Answered question

2022-02-22

Compute the limit:

\sum_{k = 1}^{\infty} \ln (1 + \frac{1}{4 k^{2}})

Lucian Robin

Beginner2022-02-23Added 4 answers

\log (1 + \frac{1}{4 k^{2}}) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} \frac{1}{4^{n + 1} k^{2 n + 2}}

Then

\sum_{k = 1}^{\infty} \log (1 + \frac{1}{4 k^{2}}) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} \frac{ζ (2 n + 2)}{4^{n + 1}}

= \sum_{n = 1}^{\infty} \frac{π^{2 n}}{2 n} \frac{B_{2 n}}{(2 n)!}

= \frac{π}{2} + \int_{0}^{1} (\sum_{n = 1}^{\infty} π^{n} t^{n - 1} \frac{B_{n}}{(n)!}) d t

= \frac{π}{2} + \int_{0}^{1} (\frac{π}{e^{π t} - 1} - \frac{1}{t}) d t

= \frac{π}{2} + \int_{0}^{π} (\frac{1}{e^{u} - 1} - \frac{1}{u}) d u

= \frac{π}{2} + \log (\frac{1 - e^{- u}}{u}) ∣_{0}^{π} = \frac{π}{2} + \log (\frac{1 - \exp (- π)}{π}) = \log (\frac{2}{π} \cdot sinh \frac{π}{2})

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