# Tag Info

3

Let's count the number of permutations with a cycle of length $\ge n/2$. It can only have one such cycle, for obvious reasons. First, pick a natural $n/2\le k\le n$. Next, pick a $k$-cycle from $S_n$, and finally the rest of the cycle decomposition can be determined by choosing any permutation on the set of elements in $\{1,\cdots,n\}$ not already present in ...

3

$$\log a_n = \sum_{k=0}^{n-1} \left(\log((3k+1)!) - \log((n+k)!\right)$$ Next use Stirling's approximation. Note e.g. that $$\sum_{k=1}^n k \log(k) \approx \int_1^n k \log(k)\; dk \approx \dfrac{n^2 \log(n)}{2} - \dfrac{n^2}{4}$$ EDIT: Ah, the Robbins numbers: OEIS sequence A005130. It seems (R.W. Gosper's approximation) $$a_n \sim ... 3 We can write$$\tau_{r+1}(n)= \sum_{d\mid n} \tau_r(d),$$which leads to your$$\sum_{n\leqslant x} \tau_{r+1}(n) = \sum_{d\leqslant x} \left\lfloor \frac{x}{d}\right\rfloor \tau_r(d),$$but we can also write$$\tau_{r+1}(n) = \sum_{d\mid n} \tau_r\left(\frac{n}{d}\right),$$and that gives us$$\begin{align} \sum_{n\leqslant x} \tau_{r+1}(n) &= ...

2

Since $f$ is real valued, we have, for $0 < t \leqslant \tau$, $$0 < \tau^2 \leqslant \tau^2 + f(\tau)^2,$$ and therefore $$0 < \int_t^\infty \frac{d\tau}{\tau^2 + f(\tau)^2} \leqslant \int_t^\infty \frac{d\tau}{\tau^2} = \frac{1}{t}.$$

2

$$\frac{k}{\log k}\to\infty\implies\frac{n^{0.1}}{(\log n)^2}=\frac1{400}\cdot\left(\frac{n^{0.0.5}}{\log(n^{0.05})}\right)^2\to\infty$$

2

In the context of branching processes, the two statements are indeed equivalent. This is due to the following facts: The value of each random variable $X(t)$ is almost surely a nonnegative integer. The events $[X(t)=0]$ are nondecreasing with respect to $t$. Thus, $P(X(t)=0)\to P(A)$ when $t\to\infty$, where $A=\bigcup\limits_t[X(t)=0]$. On the other ...

1

more basic looking, by L'hospital $$\lim_{x \rightarrow + \infty} \frac{x}{\log x} = \infty$$ Next $$\lim_{x \rightarrow + \infty} \frac{x}{(\log x)^2 } =\lim_{x \rightarrow + \infty} \frac{x}{2 \log x } = \infty$$ The general induction step is $$\lim_{x \rightarrow + \infty} \frac{x}{(\log x)^k } =\lim_{x \rightarrow + \infty} \frac{x}{k (\log ... 1 For the N(0,1) distribution we have that:$$\mathbb{P}[X>k]=\frac{1}{\sqrt{2\pi}}\int_{k}^{+\infty}e^{-x^2/2}dx =\frac{e^{-k^2/2}}{\sqrt{2\pi}}\int_{0}^{+\infty}e^{-kx}\cdot e^{-x^2}\,dx<\frac{e^{-k^2/2}}{k\sqrt{2\pi}}.\tag{1}$$You can easily adapt this argument to the N(\mu,\sigma) distribution. For an efficient continued fraction approximation ... 1 Define$$ \lambda:=\frac{\sum_{i\not\in P(i)} X_i(n)}{\sum_{i\in P(i)} X_i(n)} $$and note that \sum_{i\not\in P(i)} X_i(n)=\mathcal{O}(X(n)/t) and \sum_{i\in P(i)} X_i(n)=\Omega(X(n)/t). Your desired result requires that \lambda=o(1). But this only necessarily follows if \mathcal{O}(f)/\Omega(f)=o(1) for all functions f, which isn't true. ... 1 Basically you should just mimic the proof of Mertens' formula. We have that $\sum_{p \leq x} \log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) = \frac{4}{3} \sum_{p \leq x} \frac{1}{p} + \sum_{p \leq x}\left(\log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) - \frac{4}{3p}\right),$ and the first term is asymptotic to \frac{4}{3} \log \log x, while ... 1 Actually the relation you wanted to write down, via Merten's 3rd theorem, should be$$\prod_{p\le x}\left(1-\frac{1}{p}\right)^{4/3}\left(1+\frac{4}{3p}+\frac{C}{p^{3/2}}\right) \ll 1. Hint: we know $(1+x)^\alpha=1+\alpha x+O(x^2)$ (for $x\approx0$) and when $\prod(1-p^{-s})$ converges.

1

$\log(1-1/x)=-1/x-O(1/x^2)$ and $\pi(x)=x/\log x+O(x/\log^2x)$ so their product is $-(1/x+O(1/x^2))(x/\log x+O(x/\log^2x))=-1/\log x-O(1/\log^2x).$

1

Necessary conditions: $\liminf_{n\to\infty}f(n)\geqslant1$ and $\limsup\limits_{n\to\infty}f(n)\leqslant2$. Sufficient conditions: $\liminf\limits_{n\to\infty}f(n)\gt1$ and $\limsup\limits_{n\to\infty}f(n)\lt2$. Examples: every constant $f(n)=\alpha$, provided $1\lt\alpha\lt2$.

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