Sequences

$\lim_{n \rightarrow \infty } \sum_{0 \leqslant k \leqslant 2 n} 2^{-k} \cos \sqrt{\frac{k}{n}}$;

$\lim_{n \rightarrow \infty } \sum_{0 \leqslant k \leqslant 2 n} 2^{-n k /(n+k)}$.

Using integration by parts, prove that

Using the inequalities $W_{2 k+1}<W_{2 k}<W_{2 k-1}$, obtain Wallis's formula

Find the limit $\lim S_{n}$.

What is the nature of monotonicity of the sequence $\left\{ S_{n}\right\}$?

Is it true that $S_{n} \leqslant 2$ for every $n \in \mathbb {N}$?

$\lim \sum_{0 \leqslant k \leqslant 2 n} \frac{k}{k^{2}+n^{2}}$;

$\lim \frac{1}{n^{2}} \sum_{0 \leqslant k \leqslant n} \sqrt{k(n-k)}$;

$\lim \sum_{0 \leqslant k \leqslant 2 n} \frac{k}{k+n^{2}}$;

$\lim \sum_{n \leqslant k \leqslant 2 n} \sin \frac{\pi }{k}$;

$\lim \sum_{1 \leqslant k<n} \frac{1}{\sqrt{k(n-k)}}$;

$\lim \sum_{1<k<n} \frac{1}{(n-k) \ln k}$.

$x_{n}=\frac{n^{3 n / 2}}{n!} \prod_{1 \leqslant k \leqslant n} \sin \frac{k}{n^{3 / 2}}$;

$x_{n}=\frac{2^{n}}{\sqrt{n}} \prod_{1 \leqslant k \leqslant n} \sin \frac{\pi k}{2 n}$.

$\lim \sin (2 \pi e n!)$;

$\lim n \sin (2 \pi e n!)$;

$\lim n^{2} \sin (2 \pi e n!)$;

$\lim n^{p} \sin \left(\pi (\sqrt{2}+1)^{n}\right)$.

$x_{n} \rightarrow l_{a}=\frac{1}{2}(1+\sqrt{1+4 a})$;

$l_{a}-x_{n} \sim \frac{C_{a}}{\left(2 l_{a}\right)^{n}}$ for some $C_{a}>0$. What is $C_{2}$ equal to?

Let $p>1$ and let $\left\{ c_{n}\right\}$ be a nonnegative sequence. Set $\alpha_{n}=\sqrt[p]{c_{1}+\sqrt[p]{c_{2}+\ldots +\sqrt[p]{c_{n}}}}$. Prove that convergence of the sequence $\left\{ \alpha_{n}\right\}$ is equivalent to boundedness above of the sequence $\left\{ \frac{1}{p^{n}} \ln c_{n}\right\}$.

For which nonnegative sequences $\left\{ c_{n}\right\}$ does the sequence $\left\{ \beta_{n}\right\}$ converge, where $\beta_{n}=c_{1}+\sqrt{c_{2}+\sqrt[3]{c_{3}+\ldots +\sqrt[n]{c_{n}}}}$?

Let $c>0$, and let $\left\{ p_{n}\right\}$ be a sequence of numbers greater than one. Set $\gamma_{n}=\sqrt[p_{1}]{c+\sqrt[p_{2}]{c+\ldots +\sqrt[p_{n}]{c}}}$. Prove that convergence of the sequence $\left\{ \gamma_{n}\right\}$ is equivalent to boundedness above of the sequence $\left\{ \frac{\ln n}{p_{1} \cdot p_{2} \cdots p_{n}}\right\}$.

$3=\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+\ldots }}}}$;

$3=\sqrt{5+\sqrt{6+2 \sqrt{7+3 \sqrt{8+4 \sqrt{9+\ldots }}}}}$;

$\cos x=\sqrt{\sin^{2} \frac{x}{2}+\cos \frac{3 x}{2} \sqrt{\sin^{2} \frac{x}{4}+\cos \frac{3 x}{4} \sqrt{\sin^{2} \frac{x}{8}+\ldots }}}$ for $x \in \left[-\frac{\pi }{2}, \frac{\pi }{2}\right]$;

$9=\sqrt{3 u_{2} u_{4}+u_{4} \sqrt{3 u_{4} u_{6}+u_{6} \sqrt{3 u_{6} u_{8}+\ldots }}}$, where $\left\{ u_{n}\right\}$ is the sequence of Fibonacci numbers, i.e. $u_{1}=u_{2}=1$ and $u_{n+1}=u_{n}+u_{n-1}$ for $n \geqslant 2$;

$2 x^{2}=\sqrt{2 x T_{0} T_{1}+T_{1} \sqrt{2 x T_{1} T_{2}+T_{2} \sqrt{2 x T_{2} T_{3}+\ldots }}}$ for $x \geqslant 1$; here $\left\{ T_{n}\right\}$ is the sequence of values of the Chebyshev polynomials at the point $x$ (see problem I.2.20);

$2=\sqrt[3]{2+1 \cdot 2 \sqrt[3]{3+2 \cdot 3 \sqrt[3]{4+3 \cdot 4 \sqrt[3]{5+\ldots }}}}$.

Prove that $\overline{\lim }\left(\frac{x_{1}+x_{n+1}}{x_{n}}\right)^{n} \geqslant e$ for any positive sequence $\left\{ x_{n}\right\}$.

For any number $x \geqslant e$ exhibit a positive sequence $\left\{ x_{n}\right\}$ such that $\left(\frac{x_{1}+x_{n+1}}{x_{n}}\right)^{n} \rightarrow x$.

Let a sequence of real numbers $\left\{ x_{n}\right\}$ be such that $x_{n+m} \leqslant x_{n}+x_{m}$ for all $n, m \in \mathbb {N}$. Prove that $\frac{x_{n}}{n} \rightarrow \inf \frac{x_{m}}{m}$.

Let a sequence of complex numbers $\left\{ x_{n}\right\}$ be such that $\left|x_{n+m}-x_{n}-x_{m}\right| \leqslant A$ for all $n, m \in \mathbb {N}$. Prove that a finite limit $C=\lim \frac{x_{n}}{n}$ exists, and $\left|\frac{x_{n}}{n}-C\right| \leqslant \frac{A}{n}$.

Describe all sequences of complex numbers $\left\{ x_{n}\right\}$ satisfying the inequality $\left|x_{n+m}-x_{n}-x_{m}\right| \leqslant \frac{1}{n+m}$ for all $n, m \in \mathbb {N}$.

Let a sequence of real numbers $\left\{ x_{n}\right\}$ be such that $x_{n+1}+x_{n} \rightarrow 0$. Prove that the set of its partial limits is either infinite or contains at most two points.

Prove that if the sequence $\left\{ x_{n}+ +x_{n+1}+\ldots +x_{n+100}\right\}$ converges, then the set of partial limits of the sequence $\left\{ x_{n}\right\}$ can be represented as the union $\Delta_{1} \cup \ldots \cup \Delta_{101}$, where $\Delta_{j}$ are closed intervals (possibly degenerate or empty) in $\overline{\mathbb {R}}$.

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 $\pm 4^{k}, k \in \mathbb {Z}_{+}$?

$x_{n+1}=\left(1-\frac{1}{n}\right) x_{n}+\frac{1}{n} x_{n-1} \quad (n \in \mathbb {N})$;

$x_{n+1}=\left(1-\frac{1}{2 n}\right) x_{n}+\frac{1}{2 n} x_{n-1} \quad (n \in \mathbb {N})$.

For which values of $x_{1}$ does the sequence $\left\{ x_{n}\right\}$ diverge? What is the asymptotic behavior of $\left\{ x_{n}\right\}$?

For which $x_{1}$ does the sequence $\left\{ x_{n}\right\}$ converge? What is the limit $\lim x_{n}$?

$x_{n} \rightarrow \frac{a}{1-b}$, if $\left|b\right|<1$;

$x_{n} \rightarrow \frac{a}{1-b}$, if $\left|b\right|>1$ and $x_{1}+\sum \frac{a_{k}}{b^{k}}=0$;

$x_{n} \sim S b^{n-1}$, if $\left|b\right|>1$ and $S=x_{1}+\sum \frac{a_{k}}{b^{k}} \neq 0$.

Let $f \in C(\left[a, b\right])$, with $f(x) \in [a, b]$ for all $x \in [a, b]$. Prove that the sequence $\left\{ x_{n}\right\}$, where $x_{1} \in [a, b], x_{n+1}= \frac{1}{n}\left(f\left(x_{1}\right)+\ldots +f\left(x_{n}\right)\right)$, converges, and its limit $x^{*}$ is a fixed point of the function $f$ (i.e., $f\left(x^{*}\right)=x^{*}$).

Does an analogous statement hold in the two-dimensional case, when instead of a function continuous on an interval we consider a continuous map of some convex compact set in the plane (for example, a disk) into itself?