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How to prove that function $f(p)= frac{1}{(p^2 - 1)} sum_{q=3}^p frac{q^2-3}{q}[qtext{ is prime}]$ is less than 0.3 for all $p$?

Let's suppose that we have a function defined as follows:

$f(p) = frac{1}{(p^2 - 1)} sum_{q=3}^p frac{q^2-3}{q}[qtext{ is prime}]$

where $p$ is a prime number and $[...]$ are Iverson brackets.

Here are the first few values of $f(p)$:

$f(3) = frac{1}{(3^2 - 1)} times frac {(3^2-3)}{3}$

$f(3) = frac{1}{8} times frac{6}{3}$

$f(3) = 0.25$

$f(5) = frac{1}{(5^2 - 1)} times [ frac{(3^2-3)}{3} +frac{(5^2-3)}{5} ]$

$f(5) = frac{1}{24} times [ frac{6}{3} +frac{22}{5} ]$

$f(5) = 0.26667$

$f(7) = frac{1}{(7^2 - 1)} times [ frac{(3^2-3)}{3} +frac{(5^2-3)}{5} +frac{(7^2-3)}{7}]$

$f(7) = frac{1}{48} times [ frac{6}{3} +frac{22}{5} +frac{46}{7}]$

$f(7) = 0.273214286$

$f(11) = frac{1}{(11^2 - 1)} times [ frac{(3^2-3)}{3} +frac{(5^2-3)}{5} +frac{(7^2-3)}{7}+frac{(11^2-3)}{11}]$

$f(11) = frac{1}{120} times [ frac{6}{3} +frac{22}{5} +frac{46}{7}+frac{118}{11}]$

$f(11) = 0.198679654$

How to prove that $f(p)$ is always less than 0.3 for all prime numbers $p$?

I wrote a program to calculate $f(p)$ for the first $1000$ primes up to $p = 7,927$ and graphically it appears to approach $0$ with a maximum value of $0.2732$ at $p=7$. From the graph, it looks like it should be easy, but for some reason, I cannot figure it out.

Notice that $f(3) < f(5) < f(7)$, but $f(7) > f(11)$ so a proof by induction may not work.

$$
begin{align}
sum_{q=3}^pfrac{q^2-3}{q}
&=int_{3^-}^{p^+}frac{x^2-3}{x},mathrm{d}pi(x)\
&=left[frac{p^2-3}{p}pi(p)-2right]-int_{3^-}^{p^+}pi(x)left(1+frac3{x^2}right)mathrm{d}x\[3pt]
&=frac{p^2}{2log(p)}+frac{p^2}{4log(p)^2}+frac{p^2}{4log(p)^3}+O!left(frac{p^2}{log(p)^4}right)
end{align}
$$
Thus, we have
$$
frac1{p^2-1}sum_{q=3}^pfrac{q^2-3}{q}simfrac1{2log(p)}+frac1{4log(p)^2}+frac1{4log(p)^3}
$$
whose plot looks like

$$f(p)=frac{1}{p^2-1}sum_{q=3}^pfrac{q^2-3}{q}$$
If we use the fact that:
$$sum_{q=3}^pfrac{q^2-3}{q}=sum_{q=3}^pq-3sum_{q=3}^pfrac1q$$
Now we know that:
$$sum_{q=3}^pq=sum_{q=1}^pq-sum_{q=1}^2q=frac{p(p+1)}{2}-3$$
$$-3sum_{q=3}^pfrac1q=-3left[sum_{q=1}^pfrac1q-sum_{q=1}^2frac1qright]=-3left[ln(p)+gamma+frac{1}{2p}-frac32right]$$
If we add these together we get:
$$f(p)=frac{1}{p^2-1}left[frac{p(p+1)}{2}-3left(ln(p)+gamma+frac{1}{2p}-frac12right)right]$$
Now if you find $f'(p)$ you can find a maximum value and find what this is

The result is true for $p = 3$, $5$, and $7$, so assume that $p = 2n+1$ for $n ge 4$.
Note that all the primes $q$ occurring in the sum are odd. Thus

$$
begin{aligned}
f(p) = f(2n+1) = & frac{1}{4n^2 + 4n} sum_{q=3}^{p, text{ with $q$ prime}} frac{q^2 - 3}{q}\
< & frac{1}{4(n^2 + n)} sum_{q=3}^{p, text{ with $q$ prime}} q \
< & frac{1}{4(n^2 + n)} sum_{q = 3}^{p, text{with $q$ odd}}
q \
= & frac{1}{4(n^2 + n)} sum_{j=1}^{n} (2j+1)\
= & frac{n^2 +2n}{4(n^2 + n)} \
= & frac{3}{10} - frac{(n-4)}{20(n+1)} \
le & frac{3}{10}. end{aligned}$$

In reality, $f(p) rightarrow 0$, since

$$f(p) le frac{1}{p^2} sum_{q le p} q
le frac{1}{p^2} sum_{q le p} p
= frac{1}{p^2} cdot p pi(p) sim 1/log p.$$