Averaging of Sequences

Give an example of a bounded divergent sequence $\left\{ x_{n}\right\}$ such that $x_{n+1}-x_{n} \rightarrow 0$.

Does there exist a bounded sequence $\left\{ x_{n}\right\}$ such that $x_{n+1}-x_{n} \rightarrow 0$, but the sequence $\left\{ \frac{1}{n}\left(x_{1}+\ldots +x_{n}\right)\right\}$ has no limit?

$\sum_{1 \leqslant k \leqslant n} \frac{1}{k} \sim \ln n$;

$\sum_{1 \leqslant k \leqslant n} k^{\alpha -1} \sim \frac{n^{\alpha }}{\alpha }$;

$\sum_{k \geqslant n} \frac{1}{k^{\alpha +1}} \sim \frac{1}{\alpha n^{\alpha }}$;

$\sum_{1 \leqslant k \leqslant n} k^{\alpha -1} \ln k \sim \frac{n^{\alpha }}{\alpha } \ln n$;

$\sum_{k \geqslant n} \frac{\ln k}{k^{\alpha +1}} \sim \frac{\ln n}{\alpha n^{\alpha }}$.

$\sum_{1 \leqslant k \leqslant n} k^{\alpha k} \sim n^{\alpha n} \quad (\alpha >0)$;

$\sum_{k \geqslant n} k^{\alpha k} \sim n^{\alpha n} \quad (\alpha <0)$;

$\sum_{1 \leqslant k \leqslant n} a^{k} k!\sim a^{n} n!\quad (a>0)$;

$\sum_{1 \leqslant k \leqslant n}(k!)^{-\alpha / k} \sim \frac{e^{\alpha }}{1-\alpha } n^{1-\alpha } \quad (\alpha <1)$;

$\sum_{k \geqslant n}(k!)^{-\alpha / k} \sim \frac{e^{\alpha }}{\alpha -1} n^{1-\alpha } \quad (\alpha >1)$.

$\ln \left(n+\frac{1}{2}\right)+\gamma <1+\frac{1}{2}+\ldots +\frac{1}{n}<\ln \left(n+\frac{1}{2}\right)+\gamma +\frac{1}{6(2 n+1)^{2}}$;

$\sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n} e^{\frac{1}{12 n+1}}<n!<\sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n} e^{\frac{1}{12 n}}$.

The following inequality gives even tighter bounds for $n$!:

$\sum_{1 \leqslant k \leqslant n} k^{p}=\frac{n^{p+1}}{p+1}+\alpha_{p}+o(1)$ $(p \in (-1,0]) ;$

$\sum_{1 \leqslant k \leqslant n} k^{p}=\frac{n^{p+1}}{p+1}+\frac{n^{p}}{2}+\beta_{p}+o(1)$ $(p \in (0,1]) ;$

$\sum_{1 \leqslant k \leqslant n} k^{p}=\frac{n^{p+1}}{p+1}+\frac{n^{p}}{2}+\frac{p}{12} n^{p-1}+\gamma_{p}+o(1)$ $(p \in (1,2])$.

$\sum_{1<k \leqslant n} \frac{1}{k} \ln^{p} k=\frac{1}{p+1} \ln^{p+1} n+C_{p}+o(1) \quad (p>-1)$;

$\sum_{1<k \leqslant n} \frac{1}{k \ln k}=\ln (\ln n)+C+o(1)$;

$\sum_{1 \leqslant k \leqslant n} \frac{1}{\sqrt[k]{k!}}=e \ln n+C^{\prime }+o(1)$.

$\theta (E)=\lim \frac{k}{n_{k}}$;

for any number $\alpha \in [0,1]$ there exists a set $E \subset \mathbb {N}$ such that $\theta (E)=\alpha$.

if $x_{n} \rightarrow a \in \mathbb {R}$, then $\frac{1}{n} \sum_{k=1}^{n}\left|x_{k}-a\right| \rightarrow 0$;

if the relation $\frac{1}{n} \sum_{k=1}^{n}\left|x_{k}-a\right| \rightarrow 0$ holds for some $a \in \mathbb {R}$, then there exists a sequence $\left\{ n_{k}\right\}$ of density 1 (see Problem 2.14) such that $x_{n_{k}} \rightarrow a$;

if the sequence $\left\{ x_{n}\right\}$ is bounded and $x_{n_{k}} \rightarrow a$ for some sequence $\left\{ n_{k}\right\}$ of density 1, then $\frac{1}{n} \sum_{k=1}^{n}\left|x_{k}-a\right| \rightarrow 0$.

Show by examples that the existence of one of the limits $\lim \frac{1}{n}\left(a_{1}+\ldots +a_{n}\right)$ and $\lim \frac{1}{n}\left(a_{1}^{2}+\ldots +a_{n}^{2}\right)$ is not related to the existence of the other.

Suppose $\frac{1}{n}\left(a_{1}+\ldots +a_{n}\right) \rightarrow a$ and $\frac{1}{n}\left(a_{1}^{2}+\ldots +a_{n}^{2}\right) \rightarrow b$. Prove that $a^{2} \leqslant b \leqslant a$ and that $b$ can take any value between $a^{2}$ and $a$.

Prove that if the set $E$ has a density, then it does not exceed its periodic density.

Prove that a set with zero periodic density has zero density.

Give an example of a set with zero density whose periodic density equals one.

What is the periodic density of the set of all natural numbers representable as a sum of distinct terms of the form $\pm 4^{k}, k \in \mathbb {Z}_{+}$?