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5

We have $$e^z - 1 - z - \frac{z^2}{2} = \sum_{n = 3}^\infty \frac{z^n}{n!},$$ hence $$\biggl\lvert e^z - 1 - z - \frac{z^2}{2}\biggr\rvert \leqslant \sum_{n = 3}^\infty \frac{\lvert z\rvert^n}{n!}.$$ If we require $\lvert z\rvert < 1$, then we have $\lvert z\rvert^n < \lvert z\rvert^3$ for all $n \geqslant 3$, and that gives you an explicit $K$.


4

In fact if $g_k:[a_k,\infty) \to (0,\infty)$ is any sequence of functions such that $\lim_{x\to \infty}g_k(x) = \infty$ for all $k,$ then there exists an increasing continuous $f: [0,\infty) \to (0,\infty)$ with $\lim_{x\to \infty}f(x) = \infty$ such that $$\lim_{x\to \infty}\frac{f(x)}{g_k(x)} = 0$$ for all $k.$ Proof idea: Choose $0<b_1<b_2 < ...


4

Yes, by definition, $$f(x) \sim g(x) \quad \mbox{as} \quad x \to \infty$$ means $$\lim_{ x \to \infty} \frac{f(x)}{g(x)} = 1$$


3

Let the function $f(x)$ be given by $$f(x)=\left(1-\beta\frac{\log \log x}{\log x}\right)^{\beta}$$ where $x>1$. We wish to find the Taylor series for $f(x)$. To that end, let $g(t)$ be the function given by $$g(t)=(1-t)^{\beta}$$ and note that $f(x)=g\left(\beta\frac{\log \log x}{\log x}\right)$. Straightforward calculation shows that the ...


3

Remember that an horizontal asymptote can cross the curve (there are even examples when this happens infinitely many times). The horizontal asymptote only describes the curve's behaviour for $x\to\pm \infty$.


3

Multiply numerator and denominator by $\frac{2a}{\log x}$: $$ \frac{2a}{1-2a\frac{\log\log x}{\log x}}. $$ $a<1/2$ and $\log(x) < x$; so $2a\log\log x/\log x < 1$, and $$ f(x)=2a\sum_{n=0}^\infty \left(2a\frac{\log\log x}{\log x}\right)^n. $$ So, $f(x)\sim 2a$.


2

The infinitude of the twin primes is an open problem, so currently proving anything about the asymptotics of this function is out of reach. However, the first Hardy-Littlewood conjecture would imply that your sum is asymptotic to $$4\Pi_2 \frac{x}{\log x}$$ where $$\Pi_2=\prod_{p\geq 3} \frac{p(p-2)}{(p-1)^2}$$ is the twin prime constant.


2

For $z \rightarrow 0$ a taylor approximation is fine. I leave this part to you. For $z\rightarrow 1$ we may observe that the relevant contributions arise from the terms containing a $1-z^2$. The other contributions will only give finite corrections in this limit and can be approximated by their values at $z=1$ . Furthermore we observe that $\sinh(x)\sim ...


2

Take logs $$\log\log\log(f)=\log\log\log(2^{2^{2^n}})=n\log(2\log(2\log(2)))$$ $$\log\log(g)=\log\log(100^{100^n})=n\log (100(\log (100)))$$ Which are both linear and grow at the same rate, so to say. Now exponentiate to retrieve $f,g$.


2

Compare $f(n)$ with $G(n)=256^{256^n}>g(n)$. You arrive at $G(n) = 2^{2^{8n+3}}$, so you just need to compare $2^n$ vs. $8n+3$.


1

Yes. If we have $f(x) \in O(\sqrt{x})$, then $f(x) < k\sqrt{x}$ for some $k$ and all $x$ sufficiently large. Then we have $f(x) < \frac{k}{\sqrt{x}}x$, but since $\frac{k}{\sqrt{x}}$ gets abritarily small, we can conclude that $f(x) < cx$ or all $c$ and all $x$ sufficiently large. This is equivalent to $f(x) \in o(x)$. We can make much stronger ...


1

Two integers are coprime iff there is no prime $p$ dividing both of them, so the asymptotic probability is: $$ \frac{1}{2}\prod_{p>2}\left(1-\frac{1}{p^2}\right)=\frac{2}{3}\prod_{p}\left(1-\frac{1}{p^2}\right)=\frac{2}{3\zeta(2)}=\color{red}{\frac{4}{\pi^2}}.$$


1

This means that for some $N_0$ and some $c$ you have that whenever $n \ge N_0$: $$ 5 n - 2 \ge c (16 n + 32) $$ In this case you can take e.g. $N_0 = 100$, and set up the equation: $$ 5 N_0 - 2 = c (16 N_0 + 32) $$ which gives $c = 83 / 272$. So if $n \ge 100$: $\begin{align} (5 n - 2) - \frac{83}{272} (16 n + 32) &= \frac{2 n - 200}{17} \\ &\ge ...


1

$$\begin{eqnarray*}\sum_{j=3}^{n}\binom{n}{j}\frac{(j-1)!}{n^j}&=&\int_{0}^{+\infty}\left(\sum_{j=3}^{n}\binom{n}{j}\frac{x^{j-1}}{n^j}\right)e^{-x}\,dx\\&=&\frac{1}{2}\left(-3+\frac{1}{n}\right)+\int_{0}^{+\infty}\left(\left(1+\frac{x}{n}\right)^n-1\right)\frac{e^{-x}}{x}\,dx\end{eqnarray*}$$ hence by approximating ...


1

The approach to the asymptotics, as $n \to \infty$, of sums of the form $$ S_n:=\sum_{k=1}^n({-1})^k{n\choose k}f_k \tag1 $$ is known as the technique of Rice's integrals. The idea is to represent the sum as an integral over an appropriate complex contour then evaluating it by the residue calculus. The residue computation applied here reduces to the ...


1

Since $2^{-f(n)} > 0$, we know that $F$ is non-decreasing. Also, since $\sum 2^{-f(n)} = +\infty$, $F$ is unbounded. If $F$ attains the value $m$, then there is a smallest value $a$ with $F(a) = m$, and a smallest value $b$ with $F(b) \geqslant m+1$. Then we have by definition of $F$ $$\sum_{n = 1}^{a-1} 2^{-f(n)} < 2^m\quad\text{and}\quad \sum_{n = ...


1

$F(n)=m$ means that $$ 2^m\le\sum_{k=1}^n2^{-f(k)}\lt2^{m+1}\tag{1} $$ Let $n_m$ be the smallest $n$ so that $F(n)=m$. Then $$ \sum_{k=1}^{n_m-1}2^{-f(k)}\lt2^m\tag{2} $$ Since $f(k)\ge0$, we have $2^{-f(k)}\le1$. Therefore, since $$ \sum_{k=1}^{n_{m+1}}2^{-f(k)}\ge2^{m+1}\tag{3} $$ we have $$ \sum_{k=1}^{n_{m+1}-1}2^{-f(k)}\ge2^{m+1}-1\tag{4} $$ Therefore, ...


1

The answer is yes. The "log-star" function satisfies everything you want except for continuity, but it's easy to interpolate it to make it continuous.


1

Consider $\;\sigma:=\tanh(u)\,$ then : \begin{align} f(r)&:=\int_{0}^{\tanh(r)} \arccos\left(\frac{\sigma}{\sinh(r)\sqrt{1-\sigma^2}}\right)\cdot \frac{1}{\sqrt{\sigma^2+a^2}}\,d\sigma\\ &=\int_{0}^{r} \arccos\left(\frac{\sinh(u)}{\sinh(r)}\right) \frac{1}{\sqrt{\tanh(u)^2+a^2}}\,\frac{du}{\cosh(u)^2}\\ \end{align} Set ...



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