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Can you prove or disprove the following claim:

Let $U(n,P,Q)$ be the nth generalized Lucas number of the first kind and let $m$ be a natural number. Then, $$\sqrt{m}=1+\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \cdot (m-1)^n}{U(n,2,1-m) \cdot U(n+1,2,1-m)}$$

The SageMath cell that demonstrates this claim can be found here.

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    $\begingroup$ Not sure if it helps, but using Mathematica I found that $U(n,2,1-m)$ has the explicit formula $U(n,2,1-m)=\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k+1} m^k$. This should be easy to prove using the generating function. $\endgroup$
    – Dan Romik
    Commented Jun 18, 2021 at 6:09

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So, here we have $P=2$ and $Q=1-m$.

Notice that $$\frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = \frac{U_{n+1}(P,Q)}{U_n(P,Q)}-\frac{U_{n+2}(P,Q)}{U_{n+1}(P,Q)}.$$ By telescoping, it follows that $$\sum_{n=1}^k \frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = P - \frac{U_{k+2}(P,Q)}{U_{k+1}(P,Q)}.$$ Taking the limit over $k\to\infty$, we get $$\sum_{n=1}^\infty \frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = \frac{P-\mathrm{sgn}(P)\sqrt{P^2-4Q}}2 = 1 - \sqrt{m}.$$ QED

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