$\forall a > 0$ $\sum_{n=1}^{\infty} f(na)$ is convergent. Prove that $\int_{0}^{\infty}f(x) dx$ is convergent.

Hi can you help me solve this exercise? Thanks. Let $ f: [0;+\infty) \to \mathbb{R}$ be nonnegative and continuous function. Suppose $ \forall a > 0$ $ \sum_{n=1}^{\infty} f(na)$ is convergent. Prove that $ \int_{0}^{\infty}f(x) dx$ is convergent. I tried to solve it by using the Riemann sum, but for fixed a it doesn’t work. I have no other ideas.

Does “$\forall x\in L, \sigma(\neg x)=\neg \sigma(x)$” hold given that $\sigma(F)\equiv F$ for a CNF formula $F$ built on a set $L$ of literals?

Suppose we have a CNF formula $ F$ built on the set of literals $ L=\{x_1,\neg x_1,\cdots,x_n,\neg x_n\}$ where each variable is used in at least one clause of $ F$ . Consider a permutation $ \sigma$ of $ L$ such that $ \sigma(F)$ is logically equivalent to $ F$ i.e. $ \sigma(F)\equiv F$ .

Does it hold that $ \forall x\in L, \sigma(\neg x)=\neg \sigma(x)$ ?

I tried to find a counter example without success.

Prove that $\forall \ A_k\subset [3n]$, with $|A_k|=2n$, $\exists\ a_i,\ a_j\in A_k\ $ s.t. $\ a_i-a_j=n-1$

Let $ n\geq2$ be an integer. Prove that every subset $ A_k\subset [3n]$ , with $ |A_k|=2n$ , contains two elements $ a_i,\ a_j\ $ s.t. $ \ a_i-a_j=n-1$ , and prove that $ \nexists\ a_i,\ a_j \in A_k\ \forall k\ $ s.t. $ \ a_i – a_j =n$ .

It seems obvious that I will have to apply the Pigeonhole Principle, but I have no clear idea about how to start. Could you give me some hints? Thanks in advance!

Proove that $\forall m \in \mathbb{N}^{*},\exists n \in \mathbb{N},\forall k \geq n, p_{k+1}^m

I need to proove :

$ $ \forall m \in \mathbb{N}^{*},\exists n \in \mathbb{N},\forall k \geq n, p_{k+1}^m<\prod_{i=1}^{k}p_i$ $

I can proove this assertion using Prime number theoreme :

For fixed $ m$ we have : $ \displaystyle m \log(p_{k+1}) < \log\left( \prod_{i=1}^{k}p_i \right)$ And $ \log\left(\displaystyle \prod_{i=1}^{k}p_i \right) \sim p_k$ and $ \log(p_{k+1}) \sim \log(k+1)$ give the result.

But i need An other proof not asymptotic.

Showing $\forall x(\exists i \in I(x = x_i) \rightarrow x \in A) = \forall i \in I (x_i \in A)$

A little background information, I’m reviewing Example 2.3.1(2) from the book How To Prove It.
Basically it says that $ \forall x(\exists i \in I(x = x_i) \rightarrow x \in A)$ is the same as $ \forall i \in I (x_i \in A)$ .

What I’m not sure of is the $ (\rightarrow)$ part of the proof.

Given: $ \forall x (\exists i \in I(x = x_i) \rightarrow x \in A)$ (1)

Goal: $ \forall i \in I (x_i \in A)$ .

Scratch work

We first let i be an arbitrary element of I. Our goal thus becomes $ x_i \in A$ . Since (1) starts with $ \forall x$ , we apply universal instantiation and plug in $ x_i$ to get $ \exists i \in I(x_i = x_i) \rightarrow x_i \in A)$ .

Since $ i \in I$ and $ x_i$ is obviously equal to $ x_i$ , by modus ponens we get $ x_i \in A$ ? Is this how it is done?

Thanks in advance.

If $(\alpha,\beta)$ is the factor pair congruences of algebra $\mathbb{A},$ ia $(\forall \gamma\in ConA)\alpha\circ\gamma=\gamma\circ\beta?$

Let $ \mathbb{A}$ be an algebra such that $ ConA$ is the distributive lattice. If $ (\alpha,\beta)$ is the factor pair congruences of algebra $ \mathbb{A},$ prove that $ (\forall \gamma\in ConA)\alpha\circ\gamma=\gamma\circ\beta.$

$H \triangleleft G$, $h \in H$. $C_G(h)g \cap H \ne \emptyset, \forall g \in G$ implies $\lbrace C_G(h)g \cap H, g \in G \rbrace$ partition of $H$?

Let $ G$ be a group and $ H \triangleleft G$ . I know that $ G$ fixes the conjugacy classes of $ H$ under conjugation if and only if $ \forall h \in H, C_G(h)g \cap H \ne \emptyset, \forall g \in G$ .

Now, $ \forall h \in H$ , call $ R_g:=C_G(h)g \cap H$ . I’m arguing that, perhaps under stronger conditions that above, $ R:=\lbrace R_g, g \in G \rbrace$ is a different partition of $ H$ than the one induced by conjugacy (namely $ O=\lbrace O_h, h \in H \rbrace$ , $ O_h$ being the orbits).

Is that it?

If $\forall s\in\mathbb R^d, \forall F \in \mathcal F\ E[e^{i}I_F]=E[e^{i}]P[F]$ then $X$ is independen of $\mathcal F$

The claim in the title seems very plausible since the characteristic function “characterizes” or determines the distribution of $ X$ , but I don’t know how to derive it. There is a similar result for the characteristic functions of two Random variables, eg here, but I’m not sure if it could be deduced from that.

Any help would be appreciated!