6
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ 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
    Jun 18, 2021 at 6:09

1 Answer 1

Reset to default
10
$\begingroup$

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.