Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$\sum_{n=0}^{\infty}(-1)^n(n+1)^2\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}}=F_{k}\tag1$$
which simplifies to:
$$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k}$$
How can I evaluate the following sums to show this identity (1) is correct?
$$\sum_{n=0}^{\infty}(-1)^n\frac{n^2}{{2n \choose n}(2n+1)}=A$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{n}{{2n \choose n}(2n+1)}=B$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{1}{{2n \choose n}(2n+1)}=C$$
Best Answer
Using Wolfram for these sums, we get approximate values \begin{align*} C \approx 0.86082, \qquad B \approx -0.11649, \qquad A \approx -0.07897 \end{align*} From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define \begin{align*} g(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{\binom{2n}{n}(2n+1)}x^n \end{align*} which you can show equals \begin{align*} g(x) = 4 \frac{\sinh^{-1}(\sqrt{x}/2)}{\sqrt{x(x+4)}} \end{align*} If we define $L = x \frac{d}{dx}$ as the operator which first takes the derivative w.r.t $x$, then multiplies by $x$, we get that \begin{align*} C = (L^0 g)(1), \qquad B = (L^1 g)(1), \qquad A = (L^2 g)(1) \end{align*} or \begin{align*} C = 4 \frac{\sinh^{-1}(1/2)}{\sqrt{5}}, \qquad B = \frac{2}{5} - \frac{12 \sinh^{-1}(1/2)}{5\sqrt{5}}, \qquad A = \frac{4}{125}(7\sqrt{5}\sinh^{-1}(1/2) - 10) \end{align*}