What is $\liminf_{n\to\infty} ( \frac{p_{n}}{p_{n}+p_{n+1}})$

Question

Let $p_{n}$ a prime number and $p_{n+1}$ is the next prime.

How to calculate $\liminf_{n\to\infty} ( \frac{p_{n}}{p_{n}+p_{n+1}})$

Edited :

here is my attempt :

$\frac{p_{n}}{p_{n}+p_{n+1}}\approx \frac {n ln(n)}{p_{n}+p_{n+1}} \frac{p_{n}}{n ln(n)} $

Answer 1

In fact, your lim inf is a lim. First, observe that $$ (n+1)\ln(n+1) = n \ln(n+1) + \ln(n+1) = \\ n[\ln(n+1) - \ln(n) + \ln(n)] + \ln(n+1) = \\ n \ln(n) + n \ln\left(\frac{n+1}{n}\right) + \ln(n+1) $$ Thus (by the prime number theorem), we have $$ \lim_{n \to \infty} \frac{p_n}{p_{n+1}} = \lim_{n \to \infty} \frac{p_n}{p_{n+1}} \cdot \frac{n \ln(n)}{n \ln(n)} \cdot \frac{(n+1)\ln(n+1)}{(n+1)\ln(n+1)}\\ = \lim_{n \to \infty} \frac{p_n}{n\ln(n)} \cdot \frac{(n+1)\ln(n+1)}{p_{n+1}} \cdot \frac{n\ln(n)}{(n+1)\ln(n+1)}\\ = \lim_{n \to \infty} \frac{n\ln(n)}{(n+1)\ln(n+1)}\\ = \left[\lim_{n \to \infty} \frac{(n+1)\ln(n+1)}{n\ln(n)}\right]^{-1}\\ = \left[\lim_{n \to \infty}1 + \frac{\ln[(n+1)/n]}{\ln(n+1)} + \frac{\frac{\ln(n+1)}{\ln(n)}}{n }\right]^{-1} = 1^{-1} = 1. $$ it follows that $$ \lim_{n \to \infty} \frac{p_n}{p_n + p_{n+1}} = \lim_{n \to \infty} \frac{1}{1 + \frac{p_{n+1}}{p_n}} = \frac{1}{1+1} = \frac 12. $$

Answer 2

Siminar to Omnomnomnom's answer.

Your idea was good $$a_n= \frac{p_{n}}{p_{n}+p_{n+1}} \sim \frac{n \log (n)}{n \log (n)+(n+1) \log (n+1)}$$ Now, using Taylor expansion for large values of $n$ $$a_n=\frac{1}{2}-\frac{{\log \left({n}\right)}+1}{4 n\log(n)}+\frac{\log^2(n)+\log(n)+1}{8n^2\log^2(n) }+O\left(\frac{1}{n^2}\right)$$

Answer 3

If you're willing to use recent huge theorems in your proof, it follows from Zhang's prime gaps theorem (there is some $N$ and infinitely many $n$ with $p_{n+1} - p_n < N$).

Answer 4

Let $g_n=p_{n+1}-p_n$. Then $\frac{p_n}{p_n+p_{n+1}}=\frac{p_n}{2p_n+g_n}=\frac{1}{2+\frac{g_n}{p_n}}$. The lim inf for increasing $n$ will occur when $g_n$ is as large as possible with respect to $p_n$. However, improvements to Bertrand's postulate suggest that as $n$ becomes large, $\frac{g_n}{p_n}$ becomes very small, on the order of $\frac{1}{\ln^3 p_n}$. This is much smaller than Bertrand's postulate per se, which requires only that $\frac{g_n}{p_n}<1$

Plainly, $\lim_{n \to \infty}\frac{1}{\ln^3 p_n} \to 0$, so $\frac{1}{2+\frac{g_n}{p_n}}\approx \frac{1}{2+\frac{1}{\ln^3 p_n}} \to \frac{1}{2}$

Since the absolute limit is $\frac{1}{2}$, the lim inf is the same (as noted in a previous answer).