Can you prove or disprove the following claim:



> 
> Let $U(n,P,Q)$ be the nth [generalized Lucas number of the first kind](https://en.wikipedia.org/wiki/Lucas_sequence#Recurrence_relations) 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](https://sagecell.sagemath.org/?z=eJxdjUEKwkAQBO--Yo892Q3Z1qP4iTwgEiVKwBl1Nvt_IwQE-9RUF7SeDjzuXj7bgr4HY6kKoKUMsEhpoN9u0uFRr2M5W9XL5ISlfWKr0vzhyG2QcHt6sDBb8NHuE5iYc5Y1v7_y9gWrKx-HHydx&lang=sage&interacts=eJyLjgUAARUAuQ==).



The square root of natural number expressed by an infinite series
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


