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7

This is an incomplete answer. It only cover the case for small $a$ exactly. Part I - Exact result for small $a$ Let $\rho_c \approx 0.75487766624669$ be the unique real root of the cubic equation $\rho^3 + \rho^2 - 1 = 0$. When $a \le a_c = (1 + \rho_c)^2 \approx 3.079595623491439$, $$Q(a) = (\sqrt{a}-1)^2$$ Let $\rho = \sqrt{q}$. Given any two ...

3

We have: $$\sum_{k=0}^n {n\brack k}x^k=x(x+1)\cdots(x+n-1)$$ (See Comtet, Advanced combinatorics) and $$u_n=\frac{1}{n!} \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1}=\int_0^1\frac{t(t+1)\cdots (t+n-1)}{n!} dt$$ Now we have $\exp(-x)\geq 1-x$ for $x\geq 0$. Hence for $t\in [0,1]$ we have $$\frac{t(t+1)\cdots (t+n-1)}{n!}=t\prod_{k=1}^{n-1} ... 3 Consider the following:$$\begin{align} \frac{1}{n!}\sum_{k=1}^{n}\frac{{n\brack k}}{k+1} &\stackrel{\color{red}{[1]}}=\frac{1}{n!}\sum_{k=0}^{n}\frac{{n\brack k}}{k+1}\\ &=\frac{1}{n!}\sum_{k=0}^{n}{n\brack k}\int_{0}^{1}x^k\,\mathrm{d}x\\ &=\frac{1}{n!}\int_{0}^{1}\left(\sum_{k=0}^{n}{n\brack k}x^k\right)\,\mathrm{d}x\\ ...

2

Hint You can write $$y=\frac{\sqrt[n]{x}+1}{\sqrt[n]{x}-1}=1+\frac{2}{\sqrt[n]{x}-1}$$ So, if $x$ goes to infinity, you have $$y \simeq 1+\frac{2}{\sqrt[n]{x}}$$

2

Here is a nice simple method. If $x>1$ then $$\frac{x^4 +7x^3+5}{4x+1}<\frac{x^4+7x^4+5x^4}{4x}=\frac{13}{4}x^3$$ and $$\frac{x^4 +7x^3+5}{4x+1}>\frac{x^4}{4x+x}=\frac{1}{5}x^3\ .$$ That is, we have shown that if $x>1$ then $$\frac{1}{5}x^3<f(x)<\frac{13}{4}x^3\ .$$

2

As an expansion on the perfect comments of Daniel and Crostul. Consider the function: $$\begin{cases}f:\Bbb N\to\Bbb N \\ 2k\mapsto 3k \\ 2k+1\mapsto 2k\end{cases}$$ $$\begin{cases} g: \Bbb N\to\Bbb N \\ n\mapsto n\end{cases}$$ This sequence has $f=\Theta(g)$ by definiting, because $$|f(n)|\le 2|g(n)|$$ for every $n$, and similarly for every $n$. ...

2


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