Proposal to RMM

Let $\Re(k)>0$ and if $\displaystyle\sum_{1\leq n\leq m \leq \infty} \frac{1}{n^2m+m^2n+kmn} =\frac{(H_{k+1})^2-\psi^1(k+\alpha)}{k+\beta}+\frac{\pi^{\alpha}}{\lambda(k+\beta)} \textbf{and}$ $\displaystyle\sum_{n=0}^{\infty}\sum_{q=0}^n \frac{x^n}{(q+b)\sqrt{q+b+1}+(q+b+1)\sqrt{q+b}}=\frac{1}{\sqrt{b}(1-x)}-\Phi\left(x,\frac{1}{\alpha},b+1\right)$ where $b\in\mathbf{N}$ and $|x|<1$ then prove that $\Phi\left(\beta-\alpha , \beta,(\alpha+2\beta +\lambda)^{-1} \right)$ $= \frac{2\pi}{\sqrt5-1}+\sqrt{\phi+2}\log\left(\theta-\frac{8\theta}{4+\sqrt{10-2\sqrt5} +\sqrt{15}+\sqrt{3}}\right)+2^{-1}\sqrt{10-2\sqrt5 }\log\left(\frac{\sqrt 3\theta -1}{\theta +\sqrt 3}\right)$ and $\theta =\sqrt{\frac{8+\sqrt{10-2\sqrt 5}+\sqrt {15}+\sqrt{3}}{8-\sqrt{10-2\sqrt{5}}-\sqrt{15}-\sqrt{3}}}=\frac{8+\sqrt{10-2\sqrt 5}+\sqrt{15}+\sqrt{3}}{36-4\sqrt{5}-4\sqrt{6(5+\sqrt 5)}}$ where $\Phi(z,s,a)$ is Lerch Transcendent function , $H_k$ is the Kth Harmonic number, $\psi^1(x)$ is the trigamma function and $\phi$ is the Golden ratio.

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

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Proposal to RMM

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