Determination of all prime numbers which give integer solution of a particular summation. Determine all primes numbers $p$ such that $$p \sum_{k=0}^{n}\frac{1}{2k+1} \in N$$
for a given positive number $n$
 A: Suppose $p \sum_{k=0}^{n} \frac{1}{2k+1} \in \Bbb N$. Prime factors $q \leq n$ in the denominator of the sum can only cancel if we add a fraction with a denominator divisible by $q$. Therefore, for all primes $q \leq n$, the sum $$p\sum_{\substack{k=0\\q \mid 2k+1}}^n \frac{1}{2k+1} \in \Bbb N.$$ In particular, $\sum_{\substack{k=0\\q \mid 2k+1}}^n \frac{1}{2k+1} \in \Bbb N$ for all primes except $p$.
Now by an improvement to Betrand's postulate , for $n\geq 25$ there is always a prime $n \leq q \leq \frac{6}{5}n.$ So there are always two distinct primes $n \leq q_1 < q_2 \leq \frac{36}{25}n.$ As for any prime $q$ the next number in the sequence $(2k+1)_{k \in \Bbb N}$ that is divisible by $q$ is $3q$, for $n \geq 25$ there are always two distinct primes $q_1,q_2$, for which the only fraction in the sum with a denominator divisible by $q_1$,$q_2$ is $\frac{1}{q_1}$,$\frac{1}{q_2}$ respectively. Hence no matter what $p$ is, $p\sum_{k=0}^n \frac{1}{2k+1}$ will have a denominator that is divisible by either $q_1$ or $q_2$, a contradiction.
By testing up to $25$, the only pair $(n,p)$ for which $p \sum_{k=0}^{n} \frac{1}{2k+1} \in \Bbb N$ is $(1,3)$.
