Let $\;\displaystyle\text{Fr}_n = \sum\limits_{k=0}^n \binom{n}{k}^3\;$ be the $n^{th}$
Franel number (OEIS
A000172 ).
Method 1 (Asymptotic expansion)
By my answer to a related question, $\text{Fr}_n$ has following asymptotic expansion:
$$\text{Fr}_n = 8^n \frac{2}{\pi\sqrt{3}n}\left[
1
- \frac{1}{3n}
+ \frac{1}{27n^2}
+ \frac{1}{81n^3}
+ \frac{1}{243n^4} + \ldots
\right]$$
This immediately implies
$\displaystyle\quad \lim_{n\to\infty} \frac{\text{Fr}_{n+1}}{\text{Fr}_n}
= \lim_{n\to\infty} \frac{8^{n+1} n}{8^n (n+1)} = 8.
$
Method 2 (Poincare-Perron theorem)
In $1890s$, Franel has shown $\text{Fr}_n$ satisfy a recurrence relation of the form:
$$\text{Fr}_{n+2} = \alpha_n \text{Fr}_{n+1} + \beta_n \text{Fr}_{n}
\quad\text{ where }\quad
\begin{cases}
\alpha_n &= \frac{7 (n+1)(n+2) + 2}{(n+2)^2},\\
\beta_n &= \frac{8(n+1)^2}{(n+2)^2}
\end{cases}
$$
The key is as $n$ becomes large, both coefficients $\alpha_n$ and $\beta_n$ converge to some finite limit. The "limit" of the recurrence relation has the form
$$x_{n+2} = 7 x_{n+1} + 8 x_n$$
and its characteristic polynomial $\lambda^2 - 7 \lambda - 8 = (\lambda + 1)(\lambda - 8)$ has roots $-1$ and $8$.
Since distinct roots of the the characteristic polynomial has distinct modulus, Poincare-Perron theorem$\color{blue}{^{[1]}}$ tell us
- either $\text{Fr}_n$ vanishes for all sufficient large $n$
- or $\lim\limits_{n\to\infty} \frac{\text{Fr}_{n+1}}{\text{Fr}_n}$ converges to one of the characteristic root. i.e converges to $-1$ or $8$.
Since $\text{Fr}_n$ is positive for all $n$, it doesn't vanish for large $n$
and $\frac{\text{Fr}_{n+1}}{\text{Fr}_n}$ cannot converge to a negative number. This leaves us $\lim\limits_{n\to\infty}\frac{\text{Fr}_{n+1}}{\text{Fr}_n} = 8$ as the only possibility.
Method 3 (Binomial identities + Stirling approximations)
Let $\;\;\displaystyle C_n = \binom{n}{\lfloor\frac{n}{2}\rfloor}.\;\;$ For fixed $n > 1$ and integer $k$ such that $0 \le k \le n$, define
$$z_k = \binom{n}{k},
\quad
x_k = \begin{cases}\displaystyle \binom{n}{k-1},& k > 0\\ \\ 0,& k = 0\end{cases}
\quad\text{ and }\quad
y_k = \begin{cases}\displaystyle \binom{n}{k}, & k < n\\ \\ 0, & k = n\end{cases}
$$
Pascal identity tell us $z_k = x_k + y_k$.
Notice
$$\text{Fr}_n = \sum_{k=0}^n z_k^3\quad\text{ and }\quad
\text{Fr}_{n-1} = \sum_{k=0}^n x_k^3 = \sum_{k=0}^n y_k^3$$
We have
$$\text{Fr}_n - 8\text{Fr}_{n-1} = \sum_{k=0}^n \left( (x_k+y_k)^3 - 4 (x_k^3 + y_k^3) \right) = -3 \sum_{k=0}^n z_k (x_k - y_k)^2$$
Notice all $z_k \le C_n$, we obtain
$$|\text{Fr}_n - 8\text{Fr}_{n-1}|
\le 3 C_n \sum_{k=0}^n (x_k - y_k)^2
= 3 C_n \sum_{k=0}^n \left(2(x_k^2 + y_k^2) - z_k^2\right)
$$
It is well known
$$
\sum_{k=0}^n z_k^2 = \binom{2n}{n} = C_{2n}
\quad\text{ and }\quad
\sum_{k=0}^n x_k^2 = \sum_{k=0}^n y_k^2 = \binom{2n-2}{n-1} = \frac{n}{2(2n-1)}C_{2n}
$$
Together with a lower bound of $\text{Fr}_{n-1} \ge C_{n-1}^3$, this leads to
$$
|\text{Fr}_n - 8\text{Fr}_{n-1}| \le \frac{3}{2n-1}C_n C_{2n}
\quad\implies\quad
\left|\frac{\text{Fr}_n}{\text{Fr}_{n-1}}-8\right| \le \frac{3}{2n-1}\frac{C_n C_{2n}}{C_{n-1}^3}
$$
Using Stiring approximation, we have
$$C_n \sim 2^n \sqrt{\frac{2}{\pi n}}
\quad\implies\quad
\left|\frac{\text{Fr}_n}{\text{Fr}_{n-1}}-8\right|
\stackrel{<}{\sim} 6 \sqrt{\frac{\pi}{n}}
$$
With this, we can conclude $\;\;\lim\limits_{n\to\infty}\frac{\text{Fr}_n}{\text{Fr}_{n-1}} = 8$.
Notes
- $\color{blue}{[1]}$ I can't find any good reference for Poincare-Perron theorem online.
Please look at Chapter 8 of Saber Elaydi's book An Introduction to Difference Equations for more details.
