Let function $l(p)$ be defined as the largest prime number less than $p$.

For example: $l(7)=5$, $l(11)=7$, $l(17)=13$

Let the function $f(p)$ be defined as follows:

$$f(p) = \left(\frac{3p-3}{p^2-1}\right) + \left(\frac{1}{p-1}\right) \times \sum_{q=3}^{l(p), with \ q\ prime} \left(\frac{q+1}{q-1}\right)$$

where the summation is over prime numbers only. I.e. $q$ is always a prime.

Below are some examples of the values for $f(p)$:

$$f(3) = \frac{9-3}{9-1} + \frac{1}{3-1} \times 0$$

$$f(3) = \frac{6}{8}$$

$$f(5) = \frac{15-3}{25-1} + \frac{1}{5-1} \times \frac{3+1}{3-1}$$

$$f(5) = \frac{12}{24} + \frac{1}{4} \times 2$$

$$f(5) = 1$$

$$f(7) = \frac{21-3}{49-1} + \frac{1}{7-1} \times \left(\frac{3+1}{3-1} + \frac{5+1}{5-1}\right)$$

$$f(7) = \frac{18}{48} + \frac{1}{6} \times \left(2 + \frac {3}{2}\right)$$

$f(7) = 0.958333$

$f(11) = 0.733333$

$f(13) = 0.717063$

How to prove that $f(p)$ is always less than or equal to 1 for all prime numbers p?

Below is a graph of the values up to $p=101$. The curve appears to decline but there are exceptions, $f(31) > f(29)$ and $f(43) > f(41)$.

enter image description here


The essence of this problem is finding a good upper bound for the sum over the primes. Let $\pi(n)$ denote the prime-counting function, i.e. $\pi(n)$ is the number of primes less than or equal to $n$, and let $H_n:=\sum_{k=1}^n\tfrac1k$ denote the $n$-th harmonic sum. Then \begin{eqnarray*} \sum_{\substack{q=3\\q\ \text{prime}}}^{l(p)} \frac{q+1}{q-1} &=&\sum_{\substack{q=3\\q\ \text{prime}}}^{p-1}\left(1+\frac{2}{q-1}\right)\\ &=&\left(\pi(p-1)-1\right) +2\sum_{\substack{q=3\\q\ \text{prime}}}^{p-1}\frac{1}{q-1}\\ &\leq&\pi(p-1)-1+2\sum_{k=2}^{p-2}\frac1k\\ &=&\pi(p-1)-1+2(H_{p-2}-1)\\ &=&\pi(p-1)+2H_{p-1}-3. \end{eqnarray*} For all $n>1$ we have the well-known upper bounds $$\pi(n)\leq\frac{n}{\ln(n)}\left(1+\frac{3}{2\log(n)}\right) \qquad\text{ and }\qquad H_n\leq\ln(n+1).$$ It follows that \begin{eqnarray*} f(p) &=&\frac{3(p-1)}{p^2-1}+\frac{1}{p-1}\sum_{\substack{q=3\\q\ \text{prime}}}^{p-1} \frac{q+1}{q-1}\\ &\leq&\frac{3(p-1)}{p^2-1}+\frac{1}{p-1}(\pi(p-1)+2H_{p-2}-3)\\ &\leq&\frac{3}{p+1}+\frac{1}{p-1} \left(\frac{p-1}{\ln(p-1)}\left(1+\frac{3}{2\ln(p-1)}\right) +\ln(p-1)-3\right)\\ &=&\frac{3}{p+1}+\frac{1}{\ln(p-1)}\left(1+\frac{3}{2\ln(p-1)}\right) +\frac{\ln(p-1)}{p-1}-\frac{3}{p-1}. \end{eqnarray*} It is an exercise in basic calculus to show that the latter is strictly decreasing for $p\geq5$, and it is an exercise in basic algebra to show that the latter is less than $1$ for $p=5$.

  • $\begingroup$ Thank you for your reply! Brilliant! $\endgroup$ – temp watts Apr 2 at 20:55
  • $\begingroup$ @tempwatts My pleasure. If your question has been answered, please tick the checkmark to accept it so it doesn't remain open/unanswered, and so that it can be linked if duplicate/similar questions appear in the future. This also goes for your other questions, of course :) $\endgroup$ – Inactive - Objecting Extremism Apr 2 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.