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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Digamma and Trigamma functions

Let \psi^{(0)} and \psi^{(1)} denote the digamma and trigamma functions respectively. Prove that:

    \[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n + \frac{1}{2} \psi^{(1)}(n) \right ) = 1+ \frac{\gamma}{2} - \frac{\ln 2\pi}{2}\]

where \gamma denotes the Euler – Mascheroni constant.

Solution

We begin with the recently discovered identity:

    \begin{align*} \log \Gamma(1+x) &= \frac{\ln 2 \pi -1}{2} -\gamma \left ( x+\frac{1}{2} \right ) \\ &\quad + \frac{2x-1}{2} + \sum_{n=1}^{\infty} \left ( \psi^{(0)}(n+1) -\ln(x+n) +\frac{2x-1}{2(1+n)} \right ) \end{align*}

Letting x=1 we get that

    \[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n +\frac{1}{2n} \right ) = \frac{1+\gamma-\ln 2\pi}{2}\]

Now combining this result here we conclude the exercise.

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Trigamma series

Let \psi^{(1)} denote the trigamma function. Prove that

    \[\sum_{n=1}^{\infty} \left (\psi^{(1)} (n) - \frac{1}{n} \right )=1\]

Solution

    \begin{align*} \sum_{n=1}^{\infty} \left ( \psi^{(1)}(n) - \frac{1}{n} \right ) &= \lim_{N \rightarrow +\infty}\sum_{n=1}^{N} \left ( \psi^{(1)}(n) - \frac{1}{n} \right ) \\ &=\lim_{N \rightarrow +\infty} \left ( \log N + 1 + \gamma + \mathcal{O}\left( \frac{1}{N} \right) - \mathcal{H}_n \right ) \\ &=\lim_{N \rightarrow +\infty} \left ( \log N - \mathcal{H}_n + \mathcal{O}\left(\frac{1}{N}\right) \right ) + 1 + \gamma\\ &=-\gamma + 1 + \gamma \\ &= 1 \end{align*}

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A Riemann sum IV

Using Riemann sums prove that

    \[\lim_{n \rightarrow +\infty} \frac{\sqrt[n]{n!}}{n} = \frac{1}{e}\]

Solution

Let \displaystyle a_n = \frac{\sqrt[n]{n!}}{n}. Taking logarithms on both sides we get that

    \begin{align*} \ln a_n &= \ln \frac{\sqrt[n]{n!}}{n} \\ &=\ln \sqrt[n]{n!} - \ln n \\ &=\frac{1}{n} \ln n! - \ln n! \\ &\rightarrow -1 \end{align*}

Thus the limit follows.

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A Riemann sum III

  1. Using Riemann sums evaluate the limit

        \[\ell = \lim_{n \rightarrow +\infty} \left ( \frac{1}{n} \ln n! - \ln n \right )\]

  2. Using the above result prove that

        \[\lim_{n \rightarrow +\infty} \left ( \frac{n! e^n}{n^{n+1/2}\sqrt{2\pi}} \right )^{1/n} =1\]

Solution

  1. We have successively:

        \begin{align*} \ell &= \lim_{n \rightarrow +\infty} \left ( \frac{1}{n} \ln n! - \ln n \right ) \\ &= \lim_{n \rightarrow +\infty} \left ( \frac{1}{n} \sum_{k=1}^{n} \ln k - \ln n \right ) \\ &= \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \ln \frac{k}{n} \\ &= \int_{0}^{1} \ln x \, \mathrm{d}x \\ &= -1 \end{align*}

  2. Let \displaystyle a_n =  \left ( \frac{n! e^n}{n^{n+1/2}\sqrt{2\pi}} \right )^{1/n}. Taking logarithms on both sides we get that

        \begin{align*} \ln a_n &= \frac{1}{n} \left ( \ln n! + n \ln e - \left ( n + \frac{1}{2} \right ) \ln n - \frac{\ln 2 \pi}{2} \right ) \\ &= \frac{1}{n} \ln n! +1 - \ln n - \frac{\ln n}{2n} - \frac{\ln 2 \pi}{2n}\\ &=\left (\frac{1}{n} \ln n! - \ln n \right )+ 1 - \left ( \frac{\ln n}{n} + \frac{\ln 2\pi}{2n} \right ) \\ &\rightarrow -1 +1 -0 =0 \end{align*}

    The result follows.

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