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$\newcommand{\ts}{\tilde s}$Yes, this is true. Indeed, take any real $c\ge1/2$ and let \begin{equation} s_n(y):=\sum_{-n^2\le k\le n^2}\frac1{(ny-k)^2} \end{equation} and \begin{equation} \ts_ …

$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$For any $a=(a_1,\dots,a_n)\in(0,\infty)^n$ and any real $t\in(0,1/2)$, let $X=(X_1,\dots,X_n)$ have the Dirichlet distribution with parameter $ta$. Then $ …

$\newcommand\Ga\Gamma\newcommand\R{\mathbb R}$This answer is similar to the one linked by the OP. Indeed, let $a:=\alpha$ and $(X_{n,1},\dots,X_{n,d})$. We have $EX_{n,1}=a_1$ and $Var\,X_{n,1}\to(1-a …

We have $$\gamma \sqrt[n]{a} + (1-\gamma)\sqrt[n]{b}=\gamma a^{1/n} + (1-\gamma)b^{1/n} \\ =\gamma e^{(1/n)\ln a} + (1-\gamma)e^{(1/n)\ln b} \\ =\gamma\Big(1+\frac1n\,\ln a+O(1/n^2)\Big) + (1-\gamma)\ …

If $p>1-5d$, then $a_n(p)=0$ for all $n\ge1$ and hence $E_n(p)=0$ for all $n\ge1$. So, without loss of generality $0\le p\le1-5d$. Then, letting $E_{-1}:=1$, for all $n\ge0$ we get \begin{equation*} …

$\newcommand\ep\varepsilon\newcommand\de\delta\newcommand{\P}[1]{\overset P{\underset{#1}\longrightarrow}}$What you need is the uniform summability (in probability). Here are details: Let $Y_{l,n}:=X_ …

$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\th …

$\newcommand\ep\varepsilon$No. E.g., let $\Phi:=0$ and $$\Phi_\ep(x)=\ep e^{-|x|^2}\cos\frac{|x|}\ep$$ for all $x$, where $|x|$ is the Euclidean norm of $x$. Then $\Phi_\ep\to\Phi$ but $\nabla\Phi_\ep …

You have $$ \begin{aligned} &\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]} \\ &=\sup_{n\ge1}\|d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) \|_2 \leq 2 \sqrt{E \l …

The answer is no to both questions. E.g., suppose that $T=2$ and $$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$ Then $f_n(t)$ increases in $n$ to $f(t)=1(t<1)$, and all the other conditions o …

As suggested by Anthony Quas, the supremum in question can be rewritten as $$s_n:=\sup_{b\ge\sqrt n/A}S_n(b),$$ where $$S_n(b):=\frac1{b^3}\sum_{i=1}^n Z_i^2\,1(Z_i<b)$$ and $Z_i:=1/X_i$, so that the …

$\newcommand\ep\epsilon\newcommand{\si}{\sigma} $"I want to find a function for $N$ at which the error is $\epsilon$." This question is stated very poorly. Indeed, let $n:=N$ (there is no reason to us …

$\newcommand\la\lambda\newcommand\al\alpha\newcommand\ka\kappa\newcommand\th\theta$First of all, you copied the expression for $\al$ incorrectly. On p. 489 of the paper linked by you (the fourth displ …

Let \begin{equation*} F(x):=\sum_{n=1}^{\infty}\frac{X_n}n\,\cos nx. \tag{1}\label{1} \end{equation*} For $j=0,1,\dots$, let \begin{equation*} s_j:=\sqrt{\sum_{2^j\le n<2^{j+1}}E\Big(\frac{X_n …

$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$. Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(0,\infty]$ endowed with the st …