A hypergeometric series

Let \alpha, \beta \in \mathbb{R} such that 0<\alpha<\beta. Evaluate the series:

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right ) \cdots \left ( \alpha+n \right )}{\left (\beta+1 \right )\left ( \beta+2 \right )\cdots\left ( \beta+n \right )}\]

Solution

Lemma 1: For the \mathrm{B}, \Gamma functions , it holds that:

    \[\mathrm{B}(x, y) = \int_{0}^{1} t^{x-1} \left ( 1-t \right )^{y-1}\, \mathrm{d}t = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}\]

Lemma 2: It holds that:

    \[\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )}= \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right )\]

Proof: Simple calculations using Lemma (1) reveal the identity.

Lemma 3: Using Lemma 2 it holds that

    \[\mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right ) = \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t\]

and as a consequence

    \[\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )} = \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot   \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t\]

 

Then, successively we have that:

\begin{aligned} \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right ) \cdots \left ( \alpha+n \right )}{\left ( \beta+1 \right )\left ( \beta+2 \right )\cdots \left ( \beta+n \right )} &= \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right )_n}{\left ( \beta+1 \right )_n} \\ &=\sum_{n=0}^{\infty} \frac{\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \alpha \right )}}{\frac{\Gamma\left ( \beta+n+2 \right )}{\Gamma\left ( \beta \right )}} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right )} \cdot \sum_{n=0}^{\infty} \frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )}\\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \sum_{n=0}^{\infty} \mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right )\\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \sum_{n=0}^{\infty} \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)}\int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha} \sum_{n=0}^{\infty} t^n \, \mathrm{d}t \\ &=\frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha} \cdot \frac{\mathrm{d}t}{1-t} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha-1} \, \mathrm{d}t \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \mathrm{B} \left ( \alpha+1, \beta-\alpha \right ) \\ &=\frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \frac{\Gamma\left ( \alpha+1 \right ) \Gamma\left ( \beta-\alpha \right )}{\Gamma\left ( \beta+1 \right )} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \left( \beta -\alpha \right) \cdot \Gamma \left(\beta-\alpha \right)} \cdot \frac{\alpha \Gamma \left (\alpha \right ) \Gamma\left ( \beta - \alpha \right )}{\beta \Gamma \left ( \beta \right )} \\ &= \frac{\alpha}{\beta} \cdot \frac{1}{\beta-\alpha} \end{aligned}

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A square root limit

Let c>0. Prove that

    \[\lim _{n\rightarrow ++\infty} \left ( \frac {1+ \sqrt [n] {c}}{2} \right ) ^n= \sqrt {c}\]

Solution

It holds that

    \[x^{1/n} = 1 + \frac{\log x}{n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

Thus,

    \begin{align*} \lim_{n \rightarrow +\infty} \left ( \frac{1+\sqrt[n]{c}}{2} \right )^n &= \lim_{n \rightarrow +\infty} \left ( \frac{1+c^{1/n}}{2} \right )^n \\ &=\lim_{n \rightarrow +\infty} \left ( \frac{1}{2} +\frac{1}{2}\left ( 1+ \frac{\log c}{n} + \mathcal{O}\left ( \frac{1}{n^2} \right ) \right ) \right )^n \\ &= \lim_{n \rightarrow +\infty} \left ( 1 + \frac{\log c}{2n} + \frac{1}{2} \mathcal{O}\left ( \frac{1}{n^2} \right ) \right )^n \\ &= \lim_{n \rightarrow +\infty} \left ( 1+ \frac{\log \sqrt{c}}{n} + \frac{1}{2} \mathcal{O}\left ( \frac{1}{n^2} \right ) \right )^n \\ &= \exp \left ( \log \sqrt{c} \right ) \\ &= \sqrt{c} \end{align*}

 

Note: Similarly, it holds that \lim \limits_{n \rightarrow +\infty} \left ( 2\sqrt[n]{x}-1 \right )^n =x^2.

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Searching for the …function

Find all \mathcal{C}^1 functions f:[0, 1] \rightarrow (0, +\infty) such that \frac{f(1)}{f(0)} = e and

    \[\int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2\]

Solution

First of all we note that

    \[\int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x = \left [ \ln f(x) \right ]_0^1 = 1\]

Thus,

    \begin{align*} \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2 &\Leftrightarrow \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \\ & \quad \quad \quad \leq 2 \int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x\\ &\Leftrightarrow \int_{0}^{1} \frac{\mathrm{d}x}{f^2(x)} - 2 \int_{0}^{1} \frac{f'(x)}{f(x)} \, \mathrm{d}x + \\ & \quad \quad \quad + \int_{0}^{1} \left( f'(x) \right)^2 \, \mathrm{d}x \leq 2 \\ &\Leftrightarrow \int_{0}^{1} \left ( \frac{1}{f(x)} - f'(x) \right )^2 \, \mathrm{d}x \leq 0 \end{align*}

Since the integrand is positive it only remains that

    \begin{align*} \frac{1}{f(x)}- f'(x) =0 &\Leftrightarrow \frac{1}{f(x)} = f'(x) \\ &\Leftrightarrow f'(x) f(x) = 1 \\ &\Leftrightarrow f^2(x) = 2x+ c \end{align*}

Setting x=0 and x=1 at the last equation we have that:

    \[\begin{matrix} f^2(0) & = & c \\ f^2(1) &= &2+c \end{matrix}\Rightarrow \left ( \frac{f(0)}{f(e)} \right )^2 = \frac{c}{2+c} \Rightarrow \frac{1}{e^2} = \frac{c}{2+c} \Rightarrow c = \frac{2}{e^2-1}\]

Since f is positive we conclude that

    \begin{align*} f^2(x) = 2x+ \frac{2}{e^2-1} &\Leftrightarrow f(x) = \sqrt{2x + \frac{2}{e^2-1}} \; , \; x \in [0, 1] \end{align*}

which satisfies the given conditions.

 

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Inequality of a concave function

Let f:[0, 1] \rightarrow [0, +\infty) be a concave function. Prove that

    \[\int_0^1 x^2 f(x) \, \mathrm{d}x \leq \frac{1}{2} \int_0^1 f(x) \,\mathrm{d}x\]

Solution

Since f is concave , it holds that

    \[f\left ( ax + \left ( 1-a \right )y \right ) \geq a f(x) + \left ( 1-a \right ) f(y) \geq a f(x) \quad \text{forall} \;\; a, x, y \in [0, 1]\]

By setting y=0 and x=a we get that x f(x) \leq f \left ( x^2 \right ). Thus,

    \begin{align*} \int_{0}^{1} x^2 f(x) \, \mathrm{d}x &= \frac{1}{2} \int_{0}^{1} x f(x) 2x \, \mathrm{d}x \\ &\leq \frac{1}{2}\int_{0}^{1} f \left ( x^2 \right ) 2x \, \mathrm{d}x \\ &=\frac{1}{2} \int_{0}^{1} f \left ( x^2 \right ) \, \mathrm{d} \left ( x^2 \right ) \\ &= \frac{1}{2} \int_{0}^{1} f(x) \, \mathrm{d}x \end{align*}

and the exercise is complete.

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Convergent sequence

Let \{x_n\}_{n \in \mathbb{N}} be a sequence of positive real numbers such that

(1)   \begin{equation*} x_{n+m} \leq x_n + x_m \quad , \quad  m, n \in \mathbb{N} \end{equation*}

Prove that \left\{\dfrac{x_n}{n} \right\}_{n \in \mathbb{N}} converges.

Solution

Fix m and let n \geq m. Then, there exist k, r such that n=km+r where 0\leq r <m. Thus,

    \[\frac {x_n}{n} = \frac {x_ {km+r}}{n}\leq \frac {kx_ {m}}{n} + \frac {x_{r}}{n}\]

Letting n \rightarrow +\infty it follows that

    \[\limsup \frac {x_n}{n} \leq \frac {x_ {m}}{m} + 0\]

Since this holds forall m it follows that \displaystyle \limsup \frac {x_n}{n} \leq \liminf \frac {x_ {m}}{m} and the result follows.

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