# 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.

Note by Naren Bhandari
1 year, 3 months 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! ;)

- 1 year, 3 months ago

I started brilliant.org by the end of 2016 and now I think I learnt much from it. I hope brilliant is being helpful for you too. Glad to know you find it interesting. :)

- 1 year, 3 months ago

Yes, brilliant is quite helpful. Once I complete my 12th I will come back to this question :)

- 1 year, 3 months ago

I have shared my solution here. You may wish to see it. :)

- 1 year, 2 months ago