Proposal to RMM

Let (k)>1\Re(k)>-1 and if 1nm1n2m+m2n+kmn=(Hk+1)2ψ1(k+α)k+β+παλ(k+β)and\displaystyle\sum_{1\leq n\leq m \leq \infty} \frac{1}{n^2m+m^2n+kmn} =\frac{(H_{k+1})^2-\psi^1(k+\alpha)}{k+\beta}+\frac{\pi^{\alpha}}{\lambda(k+\beta)} \textbf{and} n=0q=0nxn(q+b)q+b+1+(q+b+1)q+b=1b(1x)Φ(x,1α,b+1)\displaystyle\sum_{n=0}^{\infty}\sum_{q=0}^n \frac{x^n}{(q+b)\sqrt{q+b+1}+(q+b+1)\sqrt{q+b}}=\frac{1}{\sqrt{b}(1-x)}-\Phi\left(x,\frac{1}{\alpha},b+1\right) where bNb\in\mathbf{N} and xR{1}x\in\mathbf{R}\setminus \left\{1\right\} then prove that Φ(βα,β,(α+2β+λ)1)\Phi\left(\beta-\alpha , \beta,(\alpha+2\beta +\lambda)^{-1} \right)=π(5(51))+2ϕ+1log(θ8θ4+1025+15+3)+21025log(3θ1θ+3) = \frac{\pi}{(5(\sqrt5-1))}+2\sqrt{\phi+1}\log\left(\theta-\frac{8\theta}{4+\sqrt{10-2\sqrt5} +\sqrt{15}+\sqrt{3}}\right)+2\sqrt{10-2\sqrt5 }\log\left(\frac{\sqrt 3\theta -1}{\theta +\sqrt 3}\right) and θ=8+1025+15+381025153=8+1025+15+3364546(5+5)\theta =\sqrt{\frac{8+\sqrt{10-2\sqrt 5}+\sqrt {15}+\sqrt{3}}{8-\sqrt{10-2\sqrt{5}}-\sqrt{15}-\sqrt{3}}}=\frac{8+\sqrt{10-2\sqrt 5}+\sqrt{15}+\sqrt{3}}{36-4\sqrt{5}-4\sqrt{6(5+\sqrt 5)}} where Φ(z,s,a)\Phi(z,s,a) is Lerch Transcendent function , HkH_k is the Kth Harmonic number, ψ1(x)\psi^1(x) is the trigamma function and ϕ\phi is the Golden ratio.

This is one of my proposed problem to Romanian Mathematical Magazine.

Note by Naren Bhandari
1 month ago

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I am in class 10 and I do not know about any of the operators mentioned in question. But it seems interesting! ;)

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