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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Trigonometric equality

Prove that in any triangle ABC it holds that

    \[\sum \sqrt{\frac{\sin A}{\sin B \sin C}} = \sqrt{\frac{2R}{r} \sum \sin A}\]

where R denotes the circumradius and r the inradius.

Solution

Using the law of sines we have that

    \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 R\]

and if we denote \mathcal{A} the area of the triangle then

    \[\frac{r \left ( a + b + c \right )}{2} = \mathcal{A} = \frac{abc}{4R}\]

Thus,

    \begin{align*} \sum \sqrt{\frac{\sin A}{\sin B \sin C}} &= \sum \sqrt{\frac{a}{2R} \cdot \frac{2R}{b} \cdot \frac{2R}{c}} \\ &=\sum \sqrt{\frac{ a}{bc} \cdot 2 R} \\ &=\sum \sqrt{\frac{a}{bc} \cdot \frac{abc}{2\mathcal{A}}} \\ &=\frac{a+b+c}{\sqrt{2 \mathcal{A}}} \\ &= \sqrt{\frac{a+b+c}{\frac{2\mathcal{A}}{a+b+c}}} \\ &= \sqrt{\frac{a+b+c}{r}} \\ &= \sqrt{\frac{1}{r} \sum a} \\ &= \sqrt{\frac{2R}{r} \sum \frac{a}{2R}} \\ &= \sqrt{\frac{2R}{r} \sum \sin A} \end{align*}

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Root inequality

Let a, b, c be three positive real numbers such that \sqrt{a} + \sqrt{b} + \sqrt{c}=1. Prove that

    \[\frac{\sqrt{a}}{a^2+2bc} + \frac{\sqrt{b}}{b^2+2ca} + \frac{\sqrt{c}}{c^2+2ab} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\]

Solution

By AM – GM we have,

    \begin{align*} \sum \frac{\sqrt{a}}{a^2+2bc} &\leq \sum \frac{\sqrt{a}}{a^2 + 2\left ( \frac{b^2+c^2}{2} \right )} \\ &= \frac{1}{a^2+b^2+c^2} \sum \sqrt{a}\\ &= \frac{1}{a^2+b^2+c^2} \end{align*}

However,

    \begin{align*} 1 &= \left (\sum \sqrt{a} \right )^2 \\ &=\left ( \sum \frac{a}{\sqrt{a}} \right )^2 \\ &\leq \left ( \sum a^2 \right ) \cdot \left ( \sum \frac{1}{a} \right ) \end{align*}

Hence \displaystyle \frac{1}{a^2+b^2+c^2} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} and the exercise is complete.

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