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?

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

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

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

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    $\begingroup$ Wow, how did you come across this monstrosity? $\endgroup$
    – YiFan Tey
    Commented Feb 4, 2019 at 22:29

3 Answers 3


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*}

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    $\begingroup$ Awesome answer! How did you come up with the generating functions and the corresponding closed form solution? Can you propose any resource for this? $\endgroup$
    – MrYouMath
    Commented Feb 4, 2019 at 23:40
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    $\begingroup$ @MrYouMath: If you write $\binom{2n}{n}=\binom{2n}{2n-n}$, then the generating function $g(x)$ is the convolution of coefficients $a_n:=\binom{2n}{n}^{-1}$ and $b_n:=(2n+1)^{-1}$, which is equivalent to the product of their individual generating functions. $\endgroup$
    – Alex R.
    Commented Feb 5, 2019 at 1:17
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    $\begingroup$ Hi @MrYouMath, Alex's method is perhaps the best way to discover the generating function. Sometimes, you can even recognize the coefficients to be similar to existing Taylor expansions. In this case, the $a_n b_n$ collectively has similar form to the coefficients in the Taylor series expansion of $\sinh^{-1}(x)$. From there, you can apply formal power series operations to try and and match the coefficients. The standard reference is generatingfunctionology by Wilf. Stanley's Enumerative Combinatorics also has an excellent presentation on generating functions / formal power series. $\endgroup$
    – Tom Chen
    Commented Feb 5, 2019 at 1:37

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.

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.

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.


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
sage: C
sage: 5*S + C

and we need to know something about hypergeometric series.


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.


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