A beautiful limit with harmonic number

Let \mathcal{H}_n denote the n – th harmonic number. Prove that

    \[\lim_{n \rightarrow  +\infty} \left(\mathcal{H}_n - \frac{1}{2^n} \sum_{k=1}^n \binom{n}{k} \mathcal{H}_k \right) = \log 2\]

Solution

We note that

    \begin{align*} \frac{1}{2^n} \sum\limits_{k=0}^n \left[ \binom{n}{k} \sum\limits_{t=1}^{k} \frac{1}{t} \right] &= \frac{1}{2^n} \sum\limits_{k=0}^n \left[ \binom{n}{k} \sum\limits_{t=1}^{k} \int_{0}^1 x^{t-1} \mathrm{d}x \right] \\ &=\frac{1}{2^n} \int_{0}^1 \sum\limits_{k=0}^n \left[ \binom{n}{k} \sum\limits_{t=1}^{k} x^{t-1} \right] \;\mathrm{d}x \\ &=\frac{1}{2^n} \int_{0}^1 \sum\limits_{k=0}^n \left[ \binom{n}{k} \cdot \frac{x^k-1}{x-1} \right]\; \mathrm{d}x \\ &=\frac{1}{2^n} \int_{0}^1 \frac{\sum\limits_{k=0}^n \binom{n}{k} x^k- \sum\limits_{k=0}^n \binom{n}{k}}{x-1} \; \mathrm{d}x \\ &=\frac{1}{2^n} \int_{0}^1 \frac{(x+1)^n- 2^n}{x-1} \; \mathrm{d}x \end{align*}

At the last integral we apply the substitution y = \frac{x+1}{2}. Thus,

    \begin{align*} \frac{1}{2^n} \int_{0}^1 \frac{(x+1)^n- 2^n}{x-1}\; \mathrm{d}x &=\int_{1/2}^{1} \frac{y^n-1}{y-1} \, \mathrm{d}y \\ &= \int_{1/2}^{1} \left ( 1+ y +y^2 + \cdots + y^{n-1} \right ) \; \mathrm{d}y \\ &= \mathcal{H}_n - \sum_{i=1}^{n} \frac{1}{i} \left ( \frac{1}{2} \right )^i \end{align*}

After all these we have that

    \begin{align*} \lim_{n \rightarrow +\infty} \left(\mathcal{H}_n - \frac{1}{2^n} \sum_{k=1}^n \binom{n}{k} \mathcal{H}_k \right) &= \lim_{n \rightarrow +\infty} \left [\cancel{ \mathcal{H}_n - \mathcal{H}_n} + \sum_{i=1}^{n} \frac{1}{i} \left ( \frac{1}{2} \right )^i \right ] \\ &= \sum_{i=1}^{\infty} \frac{1}{i} \left ( \frac{1}{2} \right )^i\\ &= \log 2 \end{align*}

proving the result.

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Zero function

Let f:\mathbb{R} \rightarrow \mathbb{R} such that f(0)=0 and \left| f'(x) \right| \leq |f(x) forall x \in \mathbb{R}. Prove that f(x)=0 forall x \in \mathbb{R}.

Solution

We note that

    \begin{align*} \left ( e^{-2x} f^2(x)\right ) ' &= 2e^{-2x}\left ( f(x)f'(x) - f^2(x) \right ) \\ &\leq 2e^{-2x}\left ( |f(x)f'(x)| - f^2(x) \right ) \\ &= 2e^{-2x}|f(x)|\left ( |f'(x)| - |f(x)| \right ) \le 0 \end{align*}

Hence e^{-2x} f^2(x) is decreasing. Therefore , e^{-2x} f^2(x) \leq 1 \cdot f^2(0) =0. The result follows.

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Limit with harmonic number

Let \mathcal{H}_n denote the n-th harmonic number. Prove that

    \[\lim_{n \rightarrow +\infty} n \left(\mathcal{H}_n -\log n -\gamma \right) = \frac{1}{2}\]

Solution

Using the well known asymptotic formula for the n – th harmonic number

    \[\mathcal{H}_n \sim \log n + \gamma + \frac{1}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

we conclude that

    \begin{align*} n \left ( \mathcal{H}_n - \log n - \gamma \right ) & \sim n \left [ \log n + \gamma + \frac{1}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right ) - \log n - \gamma \right ] \\ &\sim \frac{1}{2} + \mathcal{O} \left ( \frac{1}{n} \right ) \\ &\longrightarrow \frac{1}{2} \end{align*}

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Limit of {n \sin ( 2\pi en!)

Prove that

    \[\lim_{n \rightarrow +\infty} n \sin \left( 2\pi en! \right) = 2 \pi\]

Solution

We begin with the simple observation:

    \begin{align*} en! &= n!\sum _{k=0}^{\infty } \frac {1}{k!}\\ &= \sum _{k=0}^{n } \frac {n!}{k!}+ \sum _{k=n+1}^{\infty } \frac {n!}{k!} \\ &= \mathcal{A}_n+ \sum _{k=n+1}^{\infty } \frac {n!}{k!} \end{align*}

where \mathcal{A}_n is an integer. The last summand is of the form \displaystyle \frac{1}{n+1} + \mathcal{O} \left( \frac{1}{n^2} \right). Thus,

    \begin{align*} n \sin (2\pi e n!)&=n \sin \left [2\pi A_n+ \frac {2\pi}{n+1} + \mathcal{O} \left ( \frac{1}{n^2} \right )\right ] \\ &= n \sin \left [ \frac {2\pi}{n+1} + \mathcal{O} \left ( \frac{1}{n^2} \right ) \right ] \\ &= n \left [ \frac {2\pi}{n+1} + \mathcal{O} \left ( \frac{1}{n^3} \right )\right ] \\ &= \frac {2\pi n}{n+1} + \mathcal{O} \left ( \frac{1}{n^2} \right ) \\ &\longrightarrow 2\pi \end{align*}

and the exercise comes to an end.

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On an infinite summation

Let \{x_n\}_{n=1}^{\infty} be a sequence of real numbers. Compute:

    \[\mathcal{V} = \sum_{n=1}^{\infty} \sin^2 x_n \prod_{k=1}^{n-1} \cos^2 x_k + \prod_{n=1}^{\infty} \cos^2 x_n\]

Solution

First and foremost we set a_n = \sin^2 x_n and it is obvious that 0 \leq a_n \leq 1. We are making use of probabilistic methods. Suppose than an infinite number of coins are flipped. Let a_n be the probability that the n -th coin toss lands heads and let us consider the first time heads comes up. Then a_n \prod \limits_{k=1}^{n-1} (1 -a_k) is the probability that the first head appears in the n – th flip and \prod \limits_{n=1}^{\infty} (1-a_n) is the probability that all flips come up tails. Thus,

    \[\sum_{n=1}^{\infty} \sin^2 x_n \prod_{k=1}^{n-1} \cos^2 x_k + \prod_{n=1}^{\infty} \cos^2 x_n=1\]

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