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: \begin{eqnarray*} f(p) = \left(1 + \frac{3p-3}{p^2-1}\right) \prod_{\substack{q=3\\q\ \text{prime}}}^{l(p)} \left(1 + \frac{q+1}{q-1} \frac{1}{p-1}\right) \end{eqnarray*} Prove that $f(p)$ approaches 1 as $p$ goes to infinity.
I know that $\frac {(3p-3)}{(p^2-1)}$ and $\frac{1}{p-1}$ approach 0 as p gets large, so it makes sense that the $f(p)$ goes to 1.
Here are some example values of $f(p)$. \begin{eqnarray*} f(5) &=& \left[(1+ \frac{(3*5 – 3)}{(25-1)} \right] \left[1+ \frac{(3+1)}{(3-1)(5-1)}\right] = 2.25\\ \end{eqnarray*}
\begin{eqnarray*} f(7) &=& \left[(1+ \frac{(3*7 – 3)}{(49-1)}\right]\left[1+ \frac{(3+1)}{(3-1)(7-1)}\right] \left[1+ \frac{(5+1)}{(5-1)(7-1)}\right] = 2.292 \end{eqnarray*} $f(11) = 1.955$
$f(13) = 1.947917$
$f(17) = 1.793717$
$f(3137) = 1.154731$
Here is a plot of $f(p)$ up to $p=3137$ and it appears to approach $1$.
Let $P(n)$ be the set of primes less than $n.$
We have $\lim_{n\to \infty}1+\frac {3n-3}{n^2-1}=1.$ So it suffices to prove that $\lim_{n\to \infty}g(n)=1$ where $$g(n)=\prod_{3\le q\in P(n)}\left(1+\frac {q+1}{q-1}\frac {1}{n-1}\right).$$
Let (as usual) $\pi(n)$ be the number of primes less than $n.$ By elementary means we have $\lim_{n\to \infty}\frac {\pi(n)}{n}=0.$ Hence $$\lim_{n\to \infty}\frac {2(\pi(n)-1)}{n-1}=0.$$
Now if $3\le q<n$ then $1<1+\frac {q+1}{q-1}\frac {1}{n-1}\le 1+\frac {4}{2}\frac {1}{n-1}=1+\frac {2}{n-1}.$
And for $n\ge 4$ the number of terms in the product for $g(n)$ is $\pi(n)-1$. So for $n\ge 4$ we have $$1<g(n)\le \left(1+\frac {2}{n-1}\right)^{\pi(n)-1}=A(n)^{2(\pi(n)-1)/(n-1)}$$ $$\text { where }\quad A(n)=\left(1+\frac {2}{n-1}\right)^{(n-1)/2}. $$
Now $\lim_{n\to \infty}A(n)=e$ (See Footnote) and $\lim_{n\to \infty}2(\pi(n)-1)/(n-1)=0.$
Footnote: For the purposes of this Q it is sufficient that $A(n)\ge 1$ for $n\ge 4 $ and that $\{A(n): 4\le n\in \Bbb N\}$ is a bounded set.