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Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It …

The harmonic numbers are given by $$H_n=\sum_{k=1}^n\frac{1}{k}.$$ Numerical calculation suggests $$ \sum_{k=1}^{n}(-1)^k{n\choose k}{n+k\choose k}\sum_{i=1}^{k}\frac{1}{n+i}=(-1)^nH_n. $$ I can not g …

Calculation suggests the following identity: $$ \lim_{n\to \infty}\sum_{k=1}^{n}\frac{(-1)^k}{k}\sum_{j=1}^k\frac{1}{2j-1}=\frac{1-\sqrt{5}}{2}. $$ I have verified this identity for $n$ up to $5000$ …

Numerical calculation suggests that for prime $p\ge 5$, \begin{align*} \sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}\equiv 0\pmod{p}. \end{align*} Questi …

This is complementary to the note by Amdeberhan and Tauraso. In order to solve this problem, Amdeberhan and Tauraso [Equations (12) and (15)] established the following two important results: \begin{a …

I have verified the following double sum is always an integer for $s$ up to $1000$ via Maple. But I can not prove it. Proofs, hints, or references are all welcome. Thanks! $$\sum_{m=s}^{2s}\sum_{k=0}^{ …

Recently I came across a functional equation which always has a polynomial with integer coefficients solution. Let $$ L_n(x)=(2 x+1)^2f(x+1)-4x(x+n+1)f(x)-((2 n+1)!!)^2\prod_{i=1}^n(x+i). $$ Problem: …

The $n$-th harmonic number is defined as $$ H_n=\sum_{k=1}^{n}\frac{1}{k}, $$ and the generalized harmonic numbers are defined by $$ H_{n}^{(r)}=\sum_{k=1}^{n}\frac{1}{k^r}. $$ Recently, I have found …

Let $m$ be any positive integer. $$ P_m(x)=\sum_{i=0}^{m}\sum_{j=0}^{m}{x+j\choose j}{x-1\choose j}{j\choose i}{m\choose i}{i\choose m-j}\frac{3}{(2i-1)(2j+1)(2m-2i-1)}. $$ Question: $P_m(x)$ always h …

Let $n$ be a positive integer and $\zeta$ be a primitive $n$th root of unity. It is not hard to show that \begin{align*} \sum_{k=1}^{n-1}\frac{\zeta^k}{1-\zeta^k}=\frac{1-n}{2}. \end{align*} Since $\z …