Disclaimer: a stronger version of this question was first asked on MSE: <https://math.stackexchange.com/questions/3896547/does-p-nk1-frac1kk-p-n-whenever-0kn/3896842#3896842> and on a French math forum where one person found the counterexample $k=3$ for $n=4$. Still, the person named Ahmad proposed a sketch of proof for an asymptotic version in the link above.


So, does this inequality hold for large enough $n$?



Does $0<k<n$ imply $p_{n+k}<\left(1+\frac{1}{k}\right)^k p_{n}$ for large enough $n$?
Yes, your inequality, i.e.,


$$p\_{n+k} \lt \left(1 + \frac{1}{k}\right)^{k}p\_n\, , \; \; 0 \lt k \lt n \tag{1}\label{eq1A}$$


does hold for large enough $n$. First, an [approximation for the $n$'th prime number](https://en.wikipedia.org/wiki/Prime_number_theorem), for $n \ge 6$, is


$$n(\log(n) + \log\log(n) - 1) \lt p\_n \lt n(\log(n) + \log\log(n)) \tag{2}\label{eq2A}$$


Also, note $\left(1 + \frac{1}{k}\right)^k$ is an increasing function, with it being $2$ for $k = 1$. In addition,


$$\lim\_{k \to \infty}\left(1 + \frac{1}{k}\right)^k = e \tag{3}\label{eq3A}$$


Consider first the range $1 \le k \le \frac{n}{2}$. Then from \eqref{eq1A} and using \eqref{eq2A}, the minimum of the right side minus the maximum of the left side in the range gives


$$\begin{equation}\begin{aligned}
& 2p\_n - p\_{n+\frac{n}{2}} \\
& \gt 2n(\log(n) + \log\log(n) - 1) - \frac{3n}{2}\left(\log\left(\frac{3n}{2}\right) + \log\log\left(\frac{3n}{2}\right)\right) \\
& = n\left(2\log(n) - 2 - \frac{3}{2}\left(\log(n) + \log\left(\frac{3}{2}\right)\right) + O(\log\log(n))\right) \\
& = n\left(\frac{\log(n)}{2} - 2 - \left(\frac{3}{2}\right)\log\left(\frac{3}{2}\right) + O(\log\log(n))\right) \\
& \gt 0
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$


This shows \eqref{eq1A} holds for that range of $k$.


Next, for the remaining, slightly extended range $\frac{n}{2} \le k \le n$, \eqref{eq3A} shows for the lower bound that the multiplier of $\left(1 + \frac{2}{n}\right)^{\frac{n}{2}}$ becomes arbitrarily close to $e$ for large enough $n$, say it's $e - \epsilon$ for some arbitrarily small $\epsilon \gt 0$. Thus, as before, the minimum of the right side minus the maximum of the left side in this range gives


$$\begin{equation}\begin{aligned}
& (e - \epsilon)p\_n - p\_{2n} \\
& \gt (e - \epsilon)n(\log(n) + \log\log(n) - 1) - 2n(\log(2) + \log(n) + \log\log(2n)) \\
& = n((e - \epsilon - 2)\log(n) - (e - \epsilon) - 2\log(2) + O(\log\log(n))) \\
& \gt 0
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$


This confirms \eqref{eq1A} also holds for this other range of $k$. Thus, it holds for all $0 \lt k \lt n$ for sufficiently large $n$.


