What is $\liminf_{n\to\infty} ( \frac{p_{n}}{p_{n}+p_{n+1}})$ 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)}  $
 A: 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
$$
\lim_{n \to \infty} \frac{p_n}{p_n + p_{n+1}} = 
\lim_{n \to \infty} \frac{1}{1 + \frac{p_{n+1}}{p_n}} = \frac{1}{1+1} = \frac 12.
$$
A: 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 $n$
$$a_n=\frac{1}{2}-\frac{{\log \left({n}\right)}+1}{4
   n\log(n)}+\frac{\log^2(n)+\log(n)+1}{8n^2\log^2(n) }+O\left(\frac{1}{n^2}\right)$$
A: 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 $p_{n+1} - p_n < N$).
A: 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).
