A beautiful sum!

∑n=1∞(ψ(n)−ψ(n−12)(4n−22n−1)+ψ(n+12)−ψ(n)(4n2n))=π33+175(log⁡(1073741824)−35π2+π(ℜ{4i5log⁡(5−5+i(15−3))−2i5log⁡(5+5+i(3+15))}+65arctan⁡(53)−53)+5(log⁡(79228162514264337593543950336)log⁡(3−52)+log⁡(324518553658426726783156020576256)log⁡(3+52)+24ℜ{Li⁡2(3+i3+i5)−Li⁡2(3+i3−i5)}))\sum_{n=1}^{\infty} \left(\frac{\displaystyle \psi(n)-\psi\left(n-\frac{1}{2}\right)}{\displaystyle \binom{4n-2}{2n-1}}+\frac{\displaystyle \psi\left(n+\frac{1}{2}\right)-\psi(n)}{\displaystyle \binom{4n}{2n}}\right) = \frac{\pi}{3\sqrt{3}}+\frac{1}{75}\left(\log(1073741824)-3\sqrt{5} \pi^2+\pi\left(\Re \{{4 i \sqrt{5} \log(5- \sqrt{5} + i (\sqrt{15}- \sqrt{3}) ) - 2 i \sqrt{5} \log(5 +\sqrt{5}+ i (\sqrt{3} + \sqrt{15}))\}}+6\sqrt{5}\arctan\left(\sqrt{\frac{5}{3}}\right)-5\sqrt{3}\right)+\sqrt{5}\left(\log(79228162514264337593543950336) \log\left(\frac{3-\sqrt{5}}{2}\right) + \log(324518553658426726783156020576256)\log\left(\frac{3+\sqrt{5}}{2}\right)+24\Re \left\{\operatorname{Li}_2\left(\frac{\sqrt{3}+i}{\sqrt{3}+i\sqrt{5}}\right)-\operatorname{Li}_2\left(\frac{\sqrt{3}+i}{\sqrt{3}-i\sqrt{5}}\right)\right\}\right)\right)

Prove the above sum.

I found this sum unanswered in another mathematics forum. Can this be done without computational aid? Any ideas of solving this would be accepted.

#Calculus

Note by Aditya Kumar
4 years, 11 months ago

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Comments

It is known that ψ(n)=Hn−1−γ\psi(n) = H_{n-1} - \gamma for all n∈Nn \in \mathbb{N}. With a little work, we can also show that ψ(n+12)=2H2n−Hn−γ−ln⁡(4)\psi\left(n+\frac{1}{2}\right) = 2H_{2n} - H_n - \gamma - \ln(4).

Therefore, we can rewrite the sum as: ∑n=1∞(2Hn−1−2H2n−2+γ+ln⁡(4)(4n−22n−1)+2H2n−2Hn+1n−γ−ln⁡(4)(4n2n))\sum_{n=1}^{\infty} \left( \dfrac{2H_{n-1} - 2H_{2n-2} + \gamma + \ln(4) }{\binom{4n-2}{2n-1}} + \dfrac{2H_{2n} - 2H_n + \frac{1}{n} - \gamma - \ln(4)}{\binom{4n}{2n}} \right)

We can break these down into a few different sums, all of which obviously converge:

=∑n=1∞γ+ln⁡(4)(4n−22n−1)−∑n=1∞γ+ln⁡(4)(4n2n)−∑n=1∞2Hn−1(4n−22n−1)+∑n=1∞2H2n−2(4n−22n−1)+∑n=1∞2H2n(4n2n)−∑n=1∞2Hn(4n2n)+∑n=1∞1n(4n2n)= \sum_{n=1}^{\infty} \dfrac{\gamma + \ln(4)}{\binom{4n-2}{2n-1}} - \sum_{n=1}^{\infty} \dfrac{\gamma + \ln(4)}{\binom{4n}{2n}} - \sum_{n=1}^{\infty} \dfrac{2H_{n-1} }{\binom{4n-2}{2n-1} } + \sum_{n=1}^{\infty} \dfrac{2H_{2n-2} }{ \binom{4n-2}{2n-1}} + \sum_{n=1}^{\infty} \dfrac{2H_{2n} }{\binom{4n}{2n} } - \sum_{n=1}^{\infty} \dfrac{2H_n}{\binom{4n}{2n} } + \sum_{n=1}^{\infty} \dfrac{1}{n \binom{4n}{2n}}

We can calculate the first two sums using the following McLauren Series:

f(x)=∑k=1∞(−1)k+1x2k(2kk)=4xarcsinh(x2)(4+x2)3/2+x24+x2f(x) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1} x^{2k}}{\binom{2k}{k}} = \dfrac{4x \text{arcsinh}\left(\dfrac{x}{2}\right)}{(4+x^2)^{3/2}} + \dfrac{x^2}{4+x^2}

Then, ∑n=1∞γ+ln⁡(4)(4n−22n−1)−∑n=1∞γ+ln⁡(4)(4n2n)=(γ+ln⁡(4))f(1)\sum_{n=1}^{\infty} \dfrac{\gamma + \ln(4)}{\binom{4n-2}{2n-1}} - \sum_{n=1}^{\infty} \dfrac{\gamma + \ln(4)}{\binom{4n}{2n}} = (\gamma + \ln(4))f(1)

Of course that's just the beginning of the calculations, and it's already pretty complicated. I'm not sure how to calculate the sum of a harmonic number over a binomial coefficient.

Ariel Gershon - 4 years, 9 months ago

@Mark Hennings @Otto Bretscher

Aditya Kumar - 4 years, 11 months ago
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