A beautiful sum!

n=1(ψ(n)ψ(n12)(4n22n1)+ψ(n+12)ψ(n)(4n2n))=π33+175(log(1073741824)35π2+π({4i5log(55+i(153))2i5log(5+5+i(3+15))}+65arctan(53)53)+5(log(79228162514264337593543950336)log(352)+log(324518553658426726783156020576256)log(3+52)+24{Li2(3+i3+i5)Li2(3+i3i5)}))\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.

Note by Aditya Kumar
2 years, 11 months ago

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It is known that ψ(n)=Hn1γ\psi(n) = H_{n-1} - \gamma for all nNn \in \mathbb{N}. With a little work, we can also show that ψ(n+12)=2H2nHnγln(4)\psi\left(n+\frac{1}{2}\right) = 2H_{2n} - H_n - \gamma - \ln(4).

Therefore, we can rewrite the sum as: n=1(2Hn12H2n2+γ+ln(4)(4n22n1)+2H2n2Hn+1nγln(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=1γ+ln(4)(4n22n1)n=1γ+ln(4)(4n2n)n=12Hn1(4n22n1)+n=12H2n2(4n22n1)+n=12H2n(4n2n)n=12Hn(4n2n)+n=11n(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=1γ+ln(4)(4n22n1)n=1γ+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 - 2 years, 9 months ago

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@Mark Hennings @Otto Bretscher

Aditya Kumar - 2 years, 11 months ago

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