How to prove that the numerator is divided by 101 $1+\frac{1}{2}+\ldots+\frac{1}{100}$

Question

How to prove that: $1+\frac{1}{2}+\ldots+\frac{1}{100} = \frac{p}{q}, \gcd(p,q) = 1 \Rightarrow p \vdots 101$

I tried to allocate a numerator, but nothing happened, I tried to calculate the amount manually, but this also did not lead to success, I will be happy with any help.

Answer 1

For $1\leq n\leq 100$ let $1\leq r(n)\leq 100$ where $\frac {100!}{n}\equiv r(n)\pmod {101}.$

We have $1\leq n<n'\leq 100\implies r(n)\ne r(n').$ So $\{r(n):1\leq n\leq 100\}=\{m:1\leq m\leq 100\}.$ We have $$\sum_{n=1}^{100} \frac {1}{n}= \frac { \sum_{n=1}^{100}(100!/n)}{100!}\quad \bullet.$$ The numerator in $\bullet$ is congruent modulo $101$ to $\sum_{n=1}^{100} r(n)=\sum_{m=1}^{100} m=100\cdot 101/2=50\cdot 101,$ so the numerator in $\bullet$ is a multiple of $101.$ And the denominator $100!$ in $\bullet$ is co-prime to $101.$ So $\bullet$ in lowest terms must have a numerator divisible by $101.$

Another way is to consider this in the field $F=\Bbb Z_{101}.$ Let $S=\sum_{n=1}^{100}1/n.$ Now $F$ does not have characteristic $2$, so in $F$ we have $S=\sum_{x\in F\backslash \{0\}}(x^{-1})=\sum_{y\in F\backslash \{0\}}(y)=0.$ The implication is that in $\Bbb Z$ we have $S=A/100!$ for some $A\in \Bbb Z,$ and if $101$ does not divide $A$ then in $\Bbb Z_{101}$ we would have $S\ne 0.$

Answer 2

Hint. Note that $101$ is a prime number grater than $100$ and $$1+\frac{1}{2}+\ldots+\frac{1}{100}=\left(1+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{99}\right)+\dots+\left(\frac{1}{50}+\frac{1}{51}\right).$$

P.S. See: Wolstenholme's theorem and Proof of Wolstenholme's theorem . Actually the numerator $p$ is divided by $101^2$.

Answer 3

In ganeraly:

If p is prime number and $p>2$ and $$\frac mn=1+\frac12+\frac13+...+\frac1{p-1}$$ then $p|m$

Then $p|m$