A sum of Fibonacci numbers

Question

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?

$$\sum_{n=0}^{\infty}(-1)^n\frac{n^2}{{2n \choose n}(2n+1)}=A$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{n}{{2n \choose n}(2n+1)}=B$$

$$\sum_{n=0}^{\infty}(-1)^n\frac{1}{{2n \choose n}(2n+1)}=C$$

Answer 1

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*}

Answer 2

For the strict purpose of the question, we do not need $A,B,C$, but some linear combinations of them, that have "easy values". This answer goes mainly in this direction.

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Some words in advance.

First of all let us write the correct form of the relation: $$ \sum_{n=0}^{\infty} (-1)^n(n+1)^2 \frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}} {{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}} = \color{red}{F_{k+1}}\qquad(1) \ . $$ The simplified version of it is thus: $$ \sum_{n=0}^{\infty} (-1)^n \frac {5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}} {{2n \choose n}\cdot 2(2n+1)} = \color{red}{F_{k+1}}\qquad(2) \ . $$ This should hold for all $k$. So let us isolate in the denominator of $(2)$ an expression involving only two Fibonacci numbers, my choice being $F_k$ and $F_{k+1}$. We split for this $5n^2 = (5n^2-1)+\color{blue}1$ and form thus from $\color{blue}{1}\cdot F_k+F_{k-1}$ one $F_{k+1}$.

The other Fibonacci numbers can the be easily linearly written in terms of $F_k$ and $F_{k+1}$ (using small Fibonacci numbers as coefficiente), e.g. $F_{k+3}=2F_{k+1}+F_k$, so we have to show: $$ \sum_{n\ge0} (-1)^n \frac {(5n^2+4n+1)F_k + (5n+3)F_{k+1}} {{2n \choose n}\cdot2(2n+1)} = \color{red}{F_{k+1}}\qquad(3) \ . $$ We have thus to show that the coefficients of $F_k,F_{k+1}$ extracted from the expression in the L.H.S. are $0,2$, as in the R.H.S.

We completely avoid hypergeometric sums for this, so let us finally start the answer.

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We use the notations $$ a_n= \frac 1{\binom{2n}n}\ ,\qquad b_n= \frac n{\binom{2n}n}\ , $$ and compute simply $$ \begin{aligned} a_n+a_{n+1} &= \frac 1{\binom{2n}n} \left[1+\frac 1{\ \frac{(2n+1)(2n+2)}{n+1)(n+1)}\ }\right] = \frac {5n+3}{\binom{2n}n \cdot2(2n+1)} \ , \\ b_n+b_{n+1} &= \frac 1{\binom{2n}n} \left[n+\frac {n+1}{\ \frac{(2n+1)(2n+2)}{n+1)(n+1)}\ }\right] = \frac {5n^2+4n+1}{\binom{2n}n \cdot2(2n+1)} \ , \end{aligned} $$ and no the first relation already realizes the telescope for the coefficient of $F_{k+1}$, for short this is $\sum_{n\ge 0}(-1)^n(a_n+a_{n+1}) = a_0=1$. We simimilarly use the second relation to see that the coefficient of $F_k$ is $b_0=0$. Done.

$\square$

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If the question insists to have the values for $A,B,C$, then let us some lines. It is enough to get the value of $A$.

sage computes for it the exact value and the approximation:

sage: var('n');
sage: sum( (-1)^n / binomial(2*n, n) / (2*n+1), n, 0, oo )
4/5*sqrt(5)*log(1/2*sqrt(5) + 1/2)

sage: S = sum( (-1)^n / binomial(2*n, n)          , n, 0, oo ).n()
sage: C = sum( (-1)^n / binomial(2*n, n) / (2*n+1), n, 0, oo ).n()

sage: S
0.627836423614399
sage: C
0.860817881928008
sage: 5*S + C
4.00000000000000

and we need to know something about hypergeometric series.

Answer 3

Here is another answer, in the same spirit as that of @dan_fulea:

Both sides of $$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k+1}$$ can be considered as sequences defined by the same second order linear recurrence for the index $k$. Then, for them to be equal for all $k$, it suffices that they coincide at the first values $0$ and $1$ of the index $k$.

In other words, we just need to show that

\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n+3}{2n+1}\frac{1}{{2n \choose n}}&= F_1=1 \end{align} and \begin{align} \sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n^2+9n+4}{2n+1}\frac{1}{{2n \choose n}}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n+4}{2n+1}\frac{1}{C_n} &=F_2=1 \end{align} where we have introduced the Catalan number $C_n=\frac{1}{n+1} \binom{2n}{n}$.

But it is easy to verify, by induction, that \begin{align}\sum_{n=0}^{m}\frac{(-1)^n}{2}\frac{5n+3}{2n+1}\frac{1}{{2n \choose n}}&=\frac{{2m+2 \choose m+1}+(-1)^m}{{2m+2 \choose m+1}}\\ \sum_{n=0}^{m}\frac{(-1)^n}{2}\frac{5n+4}{2n+1}\frac{1}{C_n}&=\frac{C_{m+1}+(-1)^m}{C_{m+1}} \end{align} and we are done.