32

This follows from W. M. Schmidt's Subspace theorem, which is a deep theorem in diophantine approximations generalizing Roth's to several variables. A full account of this theorem and its proof, as well as some of its striking applications, can be found in chapter 7 of Heights in Diophantine Geometry by Bombieri and Gubler. The following result, the ...


28

Update. A research collaboration growing out of this question and some of its answers has now resulted in the following article, providing an account of the rearrangement number: A. Blass, J. Brendle, W. Brian, J. D. Hamkins, M. Hardy, and P. B. Larson, The rearrangement number, manuscript under review. Abstract. How many permutations of the natural ...


23

The sequence $a_n$ is closely related to the Wallis product $$a'_n = \prod_{i = 1}^n \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$ which converges to $\pi/2$ as $n$ goes to infinity. Namely, we have $$a'_n = a_{n + 1} \cdot \frac{2n}{2n + 1}$$. This could be proven by induction or maybe more easily by defining $b_n = a_n a_{n - 1}$ and noticing that ...


20

Banach spaces where all weakly convergent sequences are norm convergent are said to have the Schur property. A classical Theorem of Schur says that $\ell^1(I)$ has the Schur property for every set $I$. $L^1([0,1])$ does not have it. I guess that you can find this in P. Wojtaszczyk's "Banach Spaces for Analysts".


20

I have asked this question on math.stackexchange last year, and got satisfying answer. (So this construction did not come from me.) Let $(X,\mathcal O)$ be a topological space, $\mathcal F(X)$ the partialy ordered set of filters on $X$ with respect to inclusions, considered as a small category in the usual way. Given $x\in X$ and $F\in\mathcal F(X)$ let $\...


20

The product does not tend to the limit zero. For any irrational number $\alpha$ one can show that $$ \limsup_{N\to \infty} \prod_{n=1}^{N} |1- e(n\alpha)| = \infty. \tag{1} $$ (Here I use the usual notation $e(\alpha) =e^{2\pi i\alpha}$, and the product in the question is stated for $\alpha/2$ rather than $\alpha$). I'll prove a slightly weaker result;...


18

No, only if $\omega$ is a p-point. If $(A_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_k$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$. On each set $J\in \omega$ the sequence $x_n$ does not converge to 0. ...


18

By multiplying out the factor $H_n^3$, it is not too hard to see that your sum can be written as a rational linear combinations of special values of weight $4$ multiple polylogarithms. The Maple package HyperInt (by Erik Panzer) can perform simplifications with multiple polylogarithms. According to this software, $$ \sum_{n=1}^\infty \frac{H_n^3}{(n+1)2^n}=\...


16

For the purpose of recording an answer rather than just a pile of links: Michael Hardy requires that if either limit exists then so does the other and in that case then they are equal? Let's call this set $G$. Levi answered a slightly different question, namely characterizing the permutations for which if the left hand limit exists then so does the ...


16

Yes, this defines a "convergence vector space". In fact, it's probably the original motivating example for the generalization. In Fréchet's thesis he discussed L-spaces, which are essentially sequential convergence spaces: a set equipped with families of convergent sequences at each point satisfying the axioms that the constant sequence converges and ...


15

I don't know about a name, but it does have a history. Knopfmacher, Odlyzko, Pittel, Richmond, Stark [D., not H.], Szekeres, and Wormald, The asymptotic number of set partitions with unequal block sizes, available here, find that it is the residue at $z=1$ of a generating function $$G(z)=\prod_{k=1}^{\infty}\left(1-{z^k\over k!}\right)^{-1}$$ They relate it ...


14

See Erick Wong's response here. In particular, Kevin Ford proved (in more precise form) that $$ f(n) = \frac{n}{\log n} \exp\left(O(\log \log \log n)^2\right),$$ whence $f(n)/n$ tends to zero. The same consequence also follows from an earlier result of Pillai (1929), available online here.


13

looks like Somos's quadratic recurrence constant


13

No, even in the most favorable case $(X_i)_{i\geqslant 0}$ iid with $\mathbb P(X_i=1)=\mathbb P(X_i=-1)=1/2$. Denoting $F_n$ the cumulative distribution function of $n^{-1/2}S_n$, we have by symmetry $$F_{2n}(0)=\frac 12(1+\mathbb P(S_{2n}=0)).$$ Since $\mathbb P(S_{2n}=0)=\binom{2n}n2^{-2n}$, denoting $\Phi$ the cdf of the standard normal distribution, $$\...


12

Off-the-wall suggestion... Take $n$ even, I call it $2n$ now. Then asymptotically as $n \to \infty$ $$ \binom{2n}{2n-2j-1}^{-1/(2n-2j-1)} - \binom{2n}{2n-2j}^{-1/(2n-2j)} \sim \frac{1}{2n}\log \frac{2n}{2j-1} $$ and the sum $$ \frac{1}{2n}\sum_{j=1}^{n}\log\frac{2n}{2j-1} $$ is a Riemann sum for the integral $$ \frac{1}{2}\int_0^1 \log\frac{1}{t}\;dt = \...


12

Since this is community wiki, I'll feel free to share a possibly relevant anecdote; feel free to delete if you don't think this is an answer. I once had a freshman calculus student ask me if they'd be required to learn the "Greek method" for calculating derivatives. When I looked puzzled, he explained to me that the "Greek method" involved taking a limit ...


12

This is really a comment on Joel's answer, but apparently too long. Let P be the forcing which adds a permutation of $\mathbb{N}$ by finite pieces (so $P$ is forcing-equivalent to Cohen forcing). Force with a finite-support iteration of length $\omega_{1}$, using $P$ at each stage, over a ground model in which CH fails (producing, for example, the so-called ...


12

I have no idea how to approach the general problem, but here is a quick observation: A. Let $\alpha = 2\beta$ so that $\beta$ is irrational if and only if $\alpha$ is so. Define $f$ by $f(x) = \log|2\sin\pi x|$. Then $$ \log \left| \prod_{k=1}^{n} (1 - e^{\pi k i \alpha} ) \right| = \sum_{k=1}^{n} f(k\beta). $$ Now by the Riemann-integrable criterion for ...


12

Maple says $$\sum _{j=0}^{n}{\frac {{n\choose j} \left( -z \right) ^{j}}{j!}}={L}_n \left(z \right)$$ Laguerre polynomials. See https://en.wikipedia.org/wiki/Laguerre_polynomials and go down to "closed form".


11

First, we note that Stirling's series yields $4^{-k} \binom{2k}{k} = \frac{1}{\sqrt{\pi k}}e^{-1/8k + O(k^{-3})}$. Let $c = k/n$. Then $\log n!$ expands as $n \log n - n + O(\log n)$. $\log (2k)!$ expands as $2cn \log 2cn - 2cn + O(\log n)$. $\log(n-2k)!$ expands as $n(1-2c)(\log n + \log(1-2c)) - n(1-2c) + O(\log n)$. $\log \binom{n}{2k}$ expands as $-...


11

In short, $$ \begin{cases} \text{when }1\leq x & \text{series diverges when }y\le1\\ \text{when }\frac{1}{2}<x<1 & \text{series diverges when }y\leq\frac{x}{2x-1}\\ \text{when }0<x\leq\frac{1}{2} & \text{series always diverges.} \end{cases} $$ When $x>1$, the inner sum of $$\sum_{k=1}^{\infty}\frac{1}{k^{y}}\sum_{h^{x}\leq k^{y}}\...


11

See the paper "Le poisson n'a pas d'arêtes" by Thierry Bousch. This is a joke that was explained to me much later. The set is considered to resemble a fish. The French word arête means both bone and edge (of a polygon); so the English title of the paper would be "The fish has no bones/The fish has no edges". The paper proves that every point on the boundary ...


11

Here is an explicit formula for your ratio $r_n=\frac{n_n}{d_n}$: $$r_n= \frac{\sum_{k=0}^n\binom{n+k}{2k}(-x)^k} {\sum_{k=0}^n\binom{n+k+1}{2k+1}(-x)^k}.$$ Let $P_n(x)$ and $Q_n(x)$ be the numerator and denominator polynomials of $r_n$, respectively. Then both polynomials share a common recurrence; namely, $$P_{n+2}+(x-2)P_{n+1}+P_n=0 \qquad \text{and} \...


11

Such an ideal does not exist. Indeed, suppose the contrary, and let $I$ be such an ideal. The sequence $(x_n)=(1,0,1,0,\dots)$ is almost convergent, and therefore $I$-convergent, to $1/2$. So, $$\mathbb N=\{n\in\mathbb N\colon|x_n-1/2|\ge1/2\}\in I, $$ and hence $I$ is the powerset of $\mathbb N$. So, every sequence is $I$-convergent, and therefore ...


11

This appears to be an abbreviation for presque partout, meaning almost everywhere. In the article you cite, reference is made to a paper of Hunt; the MathSciNet review for Hunt's paper (MR0236019) is in French, and begins Il s'agit d'améliorations substantielles apportées au théorème de Carleson sur la convergence presque partout des séries de Fourier...


10

OK, let's try to get a few terms. We obviously have total mass $1$, so let's get some idea of how it is distributed. Clearly, what we are really interested in is $U(h)=\int_{0}^{1-h}|x^2-h^2|^\gamma$. Then $I(\gamma)=c(\gamma+1)^2\int_0^1 U(h)\log(2h)\,dh$ with some $c>0$. The only really interesting regions are $h\approx 0,x\approx 1$ and $x\approx 0, h\...


10

The standard term is Cesàro summable, named after Ernesto Cesàro. Note that a convergent sequence is also Cesàro summable (with the same limit), but the converse does not always hold. Edit. I realize that there is some confusion, thanks to the comments of Hurkyl and jeq below. Cesàro summable is usually a property attributed to a series $\sum_i b_i$. ...


10

$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$ Let $V_k:=1/\sqrt{X_k}$, $b_n:=\sqrt{n\...


10

Yes. By Proposition 2.1.10 in Lothaire, Algebraic Combinatorics on Words, if $u$ is any substring of the Fibonacci word then $$\left| \frac{\mbox{number of $1$'s in $u$}}{\mbox{length of $u$}} - \frac{1}{\phi^2} \right| \leq \frac{1}{\mbox{length of $u$}}.$$ Any length $n$ substring of any $\omega \in \Omega$ is also a length $n$ substring of the Fibonacci ...


9

I would suggest to re-write the problem as $$\lim_{n\to\infty} \sum_{k=0}^{n-1} (-1)^k \binom{n}{k}^{-1/(n-k)} = {}?\,.$$ Now if I'm not mistaken, it is true that for fixed $x$ with $|x| < 1$, we have $$\lim_{n\to\infty} \sum_{k=0}^{n-1} (-1)^k \binom{n}{k}^{-1/(n-k)} x^k = \frac{1}{1+x}$$(the coefficients have absolute value $\le 1$ and converge termwise ...


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