Prime number curiosity : $ \lim\limits_{n\to\infty}\frac{\exp(1+\sum_{i=1}^n\frac1{p_i-1})}{\prod_{i=1}^n(1+\frac1{p_i-1})}=\frac{79}{25}$ !?

Question

Let $0<n$ be a positive integer. Let $p_n$ be the $n$-th odd prime.

$$p_1 = 3,p_2=5,p_3=7,p_4=11,\ldots$$

Let

$$f(n) = \exp \left(1 + \sum_{i=1}^{n}\frac{1}{p_i-1} \right)$$

Let

$$g(n) = \prod_{i=1}^{n} \left(1+\frac{1}{p_i-1} \right)$$

Let

$$t(n) = \frac{f(n)}{g(n)}$$

Then this limit exists

$$ \lim_{n \to +\infty} t(n) = \frac{79}{25}$$

And for sufficiently large $n$ we have :

$$ \frac{79}{25} - \frac{\ln(n)+2}{n^2-1} < t(n) < \frac{79}{25} - \frac{1}{n^5}$$

How to formally prove it ?

Answer 1

Set $p_0=2$ so that $1+\sum_{i=1}^n=\sum_{i=0}^n$ and $\prod_{i=1}^n=\frac12\prod_{i=0}^n$. Then the limit is $2e^S$, where $$S=\sum_p\left[\frac1{p-1}+\log\left(1-\frac1p\right)\right]$$ (and the sum is over all primes). This can be computed very precisely, using $$S=\sum_{n=2}^\infty\frac{\varphi(n)}n\log\zeta(n)$$ with Euler's $\varphi$ and Riemann's $\zeta$, or even more efficiently, using $$S=\sum_{p\leqslant L}\left[\frac1{p-1}+\log\left(1-\frac1p\right)\right]+\sum_{n=2}^\infty\frac{\varphi(n)}n\log\zeta_{p>L}(n)$$ with $\zeta_{p>L}(s)=\zeta(s)\prod_{p\leqslant L}(1-p^{-s})$, and $L$ moderately large.

Under PARI/GP, running 2*exp(experiment(20)) (or so) after

sumprime(L,f)={my(r=0);forprime(p=2,L,r+=f(p));return(r)};
experiment(L)=sumprime(L,p->1/(p-1)+log1p(-1/p))+suminf(n=2,\
 (log(zeta(n))+sumprime(L,p->log1p(-p^(-n))))*eulerphi(n)/n);

gives $\small 2e^S=\color{red}{3.16}004224934664893010400401922739912991300144543962144985264596958636415\dots$

Note also that $S=M-\gamma+\sum_p\frac1{p(p-1)}$, where $\gamma$ is the well-known Euler–Mascheroni constant, $M$ is the less-well-known Meissel–Mertens constant, and the last sum found, say, here on OEIS.

Answer 2

This is false.

In particular, I numerically (with $30$ digits of precision) computed the product. After $555$ odd primes (prime $4021$), the total exceeds $79/25$. Since this is a strictly increasing sequence, this means that the limit is greater than $79/25$.

Running it for the first $10000$ odd primes up to $104743$ gives that this converges to $3.16004…$.