The limit is zero

Let f:(0, +\infty) \rightarrow \mathbb{R} be an integrable and uniformly continuous function. Prove that\lim \limits_{x \rightarrow + \infty} f(x) =0 .


This exercise is quite known. It first made its appearance as a Berkley exam question.

Suppose that the limit is not zero. Then let us pick a sequence x_n and an \epsilon such that foreach  n \in \mathbb{N} to both hold x_n \geq x_{n_{1}} +1 and f\left ( x_n \right )> \epsilon \quad (1). Since f is uniformly continuous there exists \delta>0 such that

\displaystyle \left | x-y \right |< \delta \implies \left | f(x) - f(y) \right | < \frac{\epsilon}{2} \quad (2)

However in the interval I_n=\left [ x_n - \frac{\delta}{2}, x_n + \frac{\delta}{2} \right ] it holds that

\displaystyle \left | \int \limits_{I_n} f(t) \, {\rm d}t \right | > \frac{\epsilon \delta }{2}

foreach n positive. The last is a consequence of (1) \; , \; (2). And of course this contradicts the fact that the function is integrable. Hence the conclusion.

Note: We can’t drop the condition of uniform continuity. If so, then the function f(x) = x \sin x^4 is a counterexample. Indeed it is continuous, integrable but the limit at infinity is not 0.

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An integral inequality

Let f:[0,1] \rightarrow \mathbb{R} be a continuous function such that

(1)   \begin{equation*} \int_0^1 f(x) \, {\rm d}x = \int_0^1 x f(x) \, {\rm d}x =1 \end{equation*}

Prove that \displaystyle \int_0^1 f^2 (x) \, {\rm d}x \geq 4 .


We note that f(x)=6x-2 satisfies all conditions. Thus:

    \begin{align*} 0 &\leq \int_{0}^{1} \left ( f(x) - 6x+2 \right )^2 \, {\rm d}x \\ &=\int_{0}^{1} f^2 (x) \, {\rm d}x - 4 \end{align*}

and thus the conclusion.

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On a determinant

Let p be a prime number and let \omega be  a primitive p-th root of unity. Define:

\mathcal{V} = \det \begin{pmatrix} 1 &1  &1  &\cdots  &1 \\  1& \omega &\omega^2  &\cdots  &\omega^{p-1} \\ 1 &\omega^2  &\left ( \omega^2 \right )^2  &\cdots  &\left ( \omega^2 \right )^{p-1} \\  \vdots&\vdots  &\vdots  &\ddots  & \vdots\\  1& \omega^{p-1} &\left ( \omega^{p-1} \right )^2  &\cdots  & \left ( \omega^{p-1} \right )^{p-1} \end{pmatrix}

Evaluate the rational number \mathcal{V}^2.


The ij -th entry of \mathcal{V} is  \omega^{(i-1)(j-1)}. Thus the ij-th entry of \mathcal{V}^2 is equal to:

\begin{aligned} \sum_{\ell} \omega^{(i-1)(\ell-1)} \omega^{(\ell-1) (j-1)} &= \sum_{\ell} \omega^{(i-1+j-1)(\ell-1)} \\ &= \left\{\begin{matrix} 0 &\text{if} & (i-1) + (j-1) \neq 0 \mod p\\ p& \text{if} & (i-1) + (j-1) = 0 \mod p \end{matrix}\right. \\ \end{aligned}

since it is known that \sum \limits_{0 \leq \ell <p} \omega^{\ell}=0 for any p-th root of unity \omega rather than 1. Thus:

\mathcal{V}^2 = \begin{pmatrix} p &0 & 0 &\cdots &0 &0 \\ 0 & 0 & 0 & \cdots & 0 & p\\ 0& 0 & 0 &\cdots & p &0 \\ \vdots& \vdots &\vdots &\ddots &\vdots &\vdots \\ 0& 0 & p& \cdots&0 & 0\\ 0& p & 0 &\cdots &0 &0 \end{pmatrix}

Thus is there is a p at the upper left corner and p ‘s along the anti diagonal in the lower right (n-1) \times (n-1) block. Thus:

\left ( \det \mathcal{V} \right )^2 = \det \left ( \mathcal{V}^2 \right )= (-1)^{(p-1)(p-2)/2} \; p^p

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A limit of a sum

Evaluate the limit

\displaystyle \lim_{n \rightarrow + \infty} e^{-n} \sum_{m=0}^{n} \frac{n^m}{m!}


This is a very well known limit and someone could argue that the value of it is 0. Unfortunately, this is not the case. We shall see that the limit equals \frac{1}{2}.

The fastest way is by making use of probabilities. Let us consider independent Poisson distributions X_1, X_2, \dots with parameter 1. Then Y_n=X_1+X_2+\cdots + X_n is Poisson with parameter n. From the central limit theorem \frac{Y_n- n}{\sqrt{n}} converges in distribution to the standard normal distribution. In particular, if N follows a standard normal distribution then

\displaystyle \lim_{n \rightarrow +\infty}\Pr(Y_n \leq n) = \lim_{n \rightarrow +\infty}\Pr\left( \frac{Y_n - n}{\sqrt{n}} \leq 0\right) = P(N \leq 0) = \frac{1}{2}

and the conclusion follows.

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An infinite product with Fibonacci

Compute the product

\displaystyle \Pi = \prod_{n=1}^{\infty} \left ( 1 + \frac{(-1)^n}{\mathcal{F}_n^2} \right )

where \mathcal{F}_n is the n -th Fibonacci sequence term.


We recall the Cassini identity that the Fibonacci sequence obeys to, hence

    \begin{align*} \Pi &= \prod_{n=1}^{\infty} \left ( 1 + \frac{(-1)^n}{\mathcal{F}_n^2} \right ) \\ &=\lim_{N \rightarrow +\infty} \prod_{n=1}^{N} \left ( 1 + \frac{(-1)^n}{\mathcal{F}_n^2} \right ) \\ &=\lim_{N \rightarrow +\infty} \prod_{n=1}^{N} \left ( \frac{\mathcal{F}_n^2 + (-1)^n}{\mathcal{F}_n^2} \right ) \\ &= \lim_{N \rightarrow +\infty} \prod_{n=1}^{N} \frac{\mathcal{F}_{n-1}}{\mathcal{F}_n} \cdot \frac{\mathcal{F}_{n+1}}{\mathcal{F}_n}\\ &= \lim_{N \rightarrow +\infty} \frac{\mathcal{F}_0 \mathcal{F}_{N+1}}{\mathcal{F}_1 \mathcal{F}_{N}} \\ & =\lim_{N \rightarrow +\infty} \frac{\mathcal{F}_{N+1}}{\mathcal{F}_{N}} \\ &= \varphi = \frac{1+\sqrt{5}}{2} \end{align*}

where \varphi is the golden ratio and its value is \varphi=\frac{1+\sqrt{5}}{2}.

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