<s>[INST] 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)}  $
 [/INST] 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
$$
\lim_{n \to \infty} \frac{p_n}{p_n + p_{n+1}} = 
\lim_{n \to \infty} \frac{1}{1 + \frac{p_{n+1}}{p_n}} = \frac{1}{1+1} = \frac 12.
$$
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