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For $n=1,2,\dots$, using the inequalities $$|w_n|\le\frac{2}{n}( |z| |w_{n-1}| + |w_{n-2}| )$$ and $$r_n:=\frac{\Gamma(n/2+1)}{n \Gamma(n/2+1/2)}\le r_1=\frac{\sqrt\pi}{2},$$ by induction on $n$ we se …

$\newcommand{\be}{\beta}\newcommand{\al}{\alpha}\newcommand{\la}{\lambda}$The integral in question is \begin{equation*} I=I_0+I_1+I_2, \tag{10}\label{10} \end{equation*} where \begin{equation*} …

At least when $n=1$, there are no nontrivial transformations of this kind. Indeed, suppose that $v(t,x)=u(\tau(t,x),\xi(t,x))$, where $u_t=u_{xx}$. Then $u_{tx}=u_{xxx}$ and $u_{tt}=u_{xxxx}$, so that …

Let $L$ denote the left-hand side of your identity \eqref{2}. Then, using the identity $$J_a(x)=\sum_{m\ge0}\frac{(-1)^m}{m!(m+a)!}(x/2)^{2m+a} \tag{$\dagger$}\label{3},$$ we get $$ \begin{split} L&=2 …

$\newcommand\ol\overline$For $x=(x_1,\dots,x_n)$, $$f_{A,B,v}(x)=\sum_{i,j,k,l,m}A_{ij}x_j\ol{B_{ik}}\ol{x_k}v_l\ol{x_l}\,\ol{v_m}x_m =\sum_{j,k,l,m}c_{j,k,l,m}x_j\ol{x_k}\,\ol{x_l}x_m,$$ where $\ol{ …

We want to upper-bound $|I_{n,k}|$, where $$I_{n,k}:=\int_0^{\pi/3} e^{-itn} (1-e^{it})^k\,dt.$$ Integrating by parts, we have $$I_{n,k}=\frac{e^{-itn}}{-in}(1-e^{it})^k\Big|_0^{\pi/3} -\int_0^{\pi/3} …

$\newcommand\od{\text{odd}}\newcommand\ev{\text{even}}$No. A counterexample is given by $$x_n=\ln n,\quad y_n=(2+(-1)^n)\ln n.$$ Then all your conditions hold, but $\varphi(a)=(1-a)/3>1-a$ for $a>1$. …

No. Indeed, make the substitution $x=u^{-1/2}$. That is, let $G_n(u):=F_n(u^{-1/2})$ and $G(u):=F(u^{-1/2})$, so that $\int_0^1\frac{F_n(x)}{x^3}\,dx=\frac12\,\int_1^\infty G_n$ and $\int_0^1\frac{F(x …

This is to complement the inequality $$|(1+it)^n-1|\ge b_1(n,t):=|e^{int}-1|,\tag{10}\label{10}$$ proved by Terry Tao for real $t$ and natural $n$, by the following inequality: $$|(1+it)^n-1|\ge b_2(n …

As in Conrad's comment, let $$f(z):=\prod_1^\infty\Big(1-\frac{z^2}{z_n^2}\Big).$$ Then for complex $z\ne0$ $$|f(z)|=\prod_1^\infty\Big|1-\frac{z^2}{z_n^2}\Big| \le\prod_1^\infty\Big(1+\frac{|z|^2}{z_ …

$\newcommand\R{\mathbb R}$Assume that the integral $$I_n:=\oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\,\frac{d\xi}{\xi^{n+1}} =\int_0^{2\pi} \frac{1-\hat{f}(e^{it})}{1-e^{it}}\,\frac{e^{it}\,i\, …

As $s\to0^+$, we have $s\Gamma(s)=\Gamma(s+1)\to1$, $x^s\to1$ unless $x=0$, and $e^{-s f(n)}\to1$ for each $n$. So, by (say) the Fatou lemma, $$ L:=\lim_{s\to0^+} \left[s \sum_{n=0}^{\infty} e^{-s f( …

In \eqref{5}, the inner sum is obviously $x(1-x)^{n-1}$. So, the limit in \eqref{5} (equal $\dfrac x{x+1}$ indeed) exists if and only if $|1-x|<2$. In \eqref{6}, using the substitution $k=K-n-j$ in th …

This is a slight modification of Alexander Kalmynin's answer. For $n\ge3$. let $z$ be a multiple root of $p_n$. Then $0=p_n'(z)=1+2z-nz^{n-1}$ and, as shown by Alexander Kalmynin, $|z|>1$. So, $n|z|^{ …

For real $k>0$, let \begin{equation} L_k:= \lim_{x\uparrow1}\Big((1-x)^{k+1} \sum_{n=1}^\infty n^k \frac{x^n}{1-x^n}\Big) =\lim_{t\downarrow0}\sum_{n=1}^\infty a_n(t), \end{equation} where \begin{eq …