TITLE: The square root of natural number expressed by an infinite series
QUESTION [6 upvotes]: 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.

REPLY [10 votes]: 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