Interesting Limit (23)

We note that $Q_n$ is defined as follows:

$\begin{aligned} Q_n & = \sqrt{\dfrac {\sum_{k=1}^n \binom nk^2}n} & \small \color{#3D99F6} \text{By Vandermonde's identity: }\binom {m+n}k = \sum_{r=0}^n \binom mr \binom n{k-r} \\ & = \sqrt{\dfrac {\binom {2n}n}n} & \small \color{#3D99F6} \text{putting }m=k=n. \end{aligned}$

Then, we have:

$\begin{aligned} \lim_{n \to \infty} \sqrt[n]{Q_n} & = \lim_{n \to \infty} \sqrt[2n]{\dfrac {\binom {2n}n}n} \\ & = \lim_{n \to \infty} \left(\frac {(2n)!}{n(n!)^2} \right)^{\frac 1{2n}} & \small \color{#3D99F6} \text{By Stirling's formula: }n! \sim \sqrt{2\pi n}\left(\frac ne\right)^n \\ & = \lim_{n \to \infty} \left(\frac {\sqrt{4\pi n}\left(\frac {2n}e\right)^{2n}}{n(2\pi n)\left(\frac ne \right)^{2n}} \right)^{\frac 1{2n}} \\ & = \lim_{n \to \infty} \left( \frac {2^{2n}}{n\sqrt{\pi n}} \right)^{\frac 1{2n}} \\ & = \lim_{n \to \infty} \frac 2{\color{#3D99F6}\pi^{\frac 1{4n}} \color{#D61F06} n^{\frac 3{4n}}} & \small \color{#3D99F6} \text{Note that }\lim_{n \to \infty} \pi^{\frac 1{4n}} = 1 \\ & = \boxed 2 & \small \color{#D61F06} \text{and }\lim_{n \to \infty} n^{\frac 3{4n}} = 1 \text{ (see note).} \end{aligned}$

Note:

$\begin{aligned} \lim_{n \to \infty} n^{\frac 3{4n}} & = \lim_{n \to \infty} \exp \left(\frac 3{4n} \ln (n) \right) & \small \color{#3D99F6} \text{where }\exp (x) = e^x \\ & = \exp \left(\color{#3D99F6} \lim_{n \to \infty} \frac {3\ln (n)}{4n} \right) & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \exp \left(\color{#3D99F6} \lim_{n \to \infty} \frac {\frac 3n}4 \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }n. \\ & = e^0 = 1 \end{aligned}$

---

References:

• Vandermonde's identity
• Stirling's formula
• L'Hôpital's rule

By the definition of the quadratic mean we $\sqrt[n]{Q_m} =\left(\dfrac{1}{n}\sum_{r=0}^{n}\left(\binom{n}{r}\right)^2\right)^{\frac{1}{2n}}=\left(\dfrac{4^n}{n}\right)^{\frac{1}{2n}}=\dfrac{2}{\sqrt[2n]{n}}$ Thus the limit is $2\lim_{n\to \infty}n^{-\frac{1}{2n}}=2\lim_{n\to \infty}\exp\left(-\dfrac{\log n}{2n}\right)=2e^{-0}=2$

Here is the proof for the first part