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Limit of geometric mean of binomial coefficients

Let G_n denote the geometric mean of the binomial coefficients

    \[\binom{n}{0}, \; \binom{n}{1}, \; \binom{n}{2}, \; \cdots, \; \binom{n}{n}\]

Prove that \lim \limits_{n \rightarrow +\infty} \sqrt[n]{G_n}=\sqrt{e}.

Solution

We note that

    \begin{align*} \sqrt[n]{G_n}&=\exp\left(\frac{1}{n(n+1)}\sum_{k=0}^n\left(\ln n!-\ln k!-\ln((n-k)!)\right)\right)\\ &=\exp\left(\frac{1}{n}\sum_{k=1}^n\ln k-\frac{2}{n(n+1)}\sum_{k=1}^n\sum_{m=1}^k\ln m\right) \end{align*}

On the other hand the following lemma holds:

Lemma: Let f:[M, N] \rightarrow \mathbb{R} be a monotonic function. It holds that

    \[\left|\sum_{k=M}^Nf(k)-\int_M^Nf(x)\,\mathrm{d}x\right|\leq\max\{|f(M)|,|f(N)|\}\]

Proof: Due to monotony it holds that f(k+1)\leq \int_k^{k+1}f(x)\,\mathrm{d}x \leq f(k)for k=M,\ldots,N-1. Hence summing over all these values of k we get that

    \[-f(M)\leq\int_M^Nf(x)\,\mathrm{d}x-\sum_{k=M}^Nf(k)\leq -f(N)\]

The result follows.

Applying the above to f(x)=\ln x on [1, k] we get that:

    \[\left|\int_{1}^{k}\ln(x)\,dx-\sum_{m=1}^k\ln m\right|\leq \ln k\]

Thus,

(1)   \begin{equation*} \sum_{m=1}^k\ln m=\int_{1}^{k}\ln x\,\mathrm{d}x+\mathcal O(\ln k)=k\ln k-k+\mathcal{O}\left(\max\{1,\ln k\}\right) \end{equation*}

Similarly, applying the above to f(x)=x\ln x on [1, n] we get that:

(2)   \begin{equation*} \sum_{k=1}^nk\ln k=\int_{1}^{n}x\ln x\,\mathrm{d}x+\mathcal O(n\ln n)=\frac{n^2\ln n}{2}-\frac{n^2}{4}+\mathcal O(n\ln n) \end{equation*}

The result follows.

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