4 votes

How to find the coefficient of $x^k$ in the expression $\prod_{p=2}^n (1+xp) $

From representation $$\prod_{p=2}^n (1+xp) = (-x)^{n+1} (-1/x)_{n+1} (1+x)^{-1}= \sum_{i\geq 0} s_1(n+1,i) (-x)^{n+1-i}\cdot \sum_{j\geq0} (-x)^j, $$ it follows that the coefficient of $x^{k-1}$ in ...
Max Alekseyev's user avatar
3 votes
Accepted

Boundedness from convergence of some demeaned sequence

While writing this answer it appears to me that I discovered some sort of a tautology, basically because $a_n$ are uniquely determined by $b_n$. Specifically, by simple algebra we have $\bar{a}_n - \...
Aleksei Kulikov's user avatar
3 votes
Accepted

Example of a conditionally convergent series $\sum_{n=1}^\infty b_n$ such that $n^2(b_n-b_{n+1})$ is bounded

The answer is yes, such an example exists. For instance, let $b_1=0$ and $b_n=(-1)^j h_j(n)$ for $j=1,2,\dots$ and $n\in N_j:=\{2^j,\dots,2^{j+1}-1\}$, where $$h_j(n):=c_j\frac{\min(n-2^j,2^{j+1}-n)}{...
Iosif Pinelis's user avatar
2 votes

Number of Salem–Spencer subsets of $\{1,2,3,\dots ,n\}$

Let $r_3(N)$ denote the maximum cardinality of a $3$-AP subset of $\{1,\dots,N\}$. It clear that $T(N)\ge 2^{r_3(N)}$. Meanwhile, it was proved by Balogh, Liu, and Sharifzadeh that we have $T(N) \le 2^...
Zach Hunter's user avatar
  • 3,393
1 vote
Accepted

Sequence with two restrictions

Yes, that works. Whatever has happened before (for $n<N_1$, say), we can always make the ratio from (2) large at some later point $N_2$ by taking $a_n=\epsilon$ very small on a very long subsequent ...
Christian Remling's user avatar

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