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 $$\sum_{n=1}^\infty \frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = \frac{P-\sqrt{P^2-4Q}}2 = 1 - \sqrt{m}.$$ QED

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

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 $$\sum_{n=1}^\infty \frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = 1 - \sqrt{m}.$$ QED

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 $$\sum_{n=1}^\infty \frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = \frac{P-\sqrt{P^2-4Q}}2 = 1 - \sqrt{m}.$$ QED