Given a prime number $p\_0$, by Bertrand's postulate we know that
\begin{gather}
p\_1\ge\frac{p\_0}{2}\\
p\_2\ge\frac{p\_1}{2}\ge\frac{p\_0}{2^2}\\
\vdots\\
p\_k\ge\frac{p\_0}{2^k}
\end{gather}
where $p\_1,p\_2,\dots$ are prime numbers immediately preceding $p\_0$.



> 
> **Question.** Is there a similar nice upper bound for the $p\_i$?
> 
> 
> 


A trivial upper bound would be $p\_i\le p\_0-2i$, but that is trivial and bad in the sense that for large $i$ this doesn't give a good approximation (since one or two primes may differ by $2$, but $10$ consecutive primes can't all differ by $2$, there must be something better). I am not claiming that the above result due to Bertrand gives a good approximation, but that at least gives exponential bounds, unlike in this case.



What is a non-trivial upper bound on the $k$th prime below a given prime $p$?
There is no good upper bound for $p\_k$ in the following sense: for every $k$, there exists a constant $c\_k>0$ such that $p\_k>p\_0-c\_k$ holds for infinitely many primes $p\_0$. This was proved by [Maynard (2013)](https://arxiv.org/abs/1311.4600). You will find a concrete value for $c\_k$ in Maynard's work, which was decreased by [Polymath8b (2014)](https://arxiv.org/abs/1407.4897). The best conjectured value would follow from the [Hardy-Littlewood prime tuple conjecture](https://en.wikipedia.org/wiki/Prime_k-tuple).


Note also that the crude lower bound $p\_k>p\_0/2^k$ can be improved greatly. For example, Baker-Harman-Pintz (2000) proved that, for $x$ sufficiently large, there is always a prime in the interval $[x-x^{21/40},x]$. In particular, we have $p\_1>p\_0-p\_0^{21/40}$ for all but finitely many primes $p\_0$. This is still far from being optimal. The Riemann Hypothesis implies that the exponent can be replaced by any number exceeding $1/2$. Further, a [conjecture of Cramér](https://en.wikipedia.org/wiki/Cram%C3%A9r%27s_conjecture) states that $x^{21/40}$ can be replaced by $2(\log x)^2$.


