Is the base of natural logarithms $e$ efficiently computable?
It is not hard to check (for example using a Taylor expansion of $e^x$ at $x=0$) that for all $n \geq 0$,
$$
\sum_{k=0}^n \frac{1}{k!} &lt; e &lt; \sum_{k=0}^n \frac{1}{k!} + \frac{e}{(n+1)!} &lt; \sum_{k=0}^n \frac{1}{k!} + \frac{3}{(n+1)!}.
$$
Moreover,
$$
\sum_{k=0}^n \frac{1}{k!} = \frac{\sum_{k=0}^n \prod_{i=k+1}^n i}{n!}.
$$
This formula gives an algorithm, running in time $\mathit{poly}(n)$, for computing $\sum_{k=0}^n \frac{1}{k!}$. This quantity satisfies your requirements if $\frac{3}{(n+1)!} \leq \frac{1}{C}$, that is, if $(n+1)! \geq 3C$. Since $(n+1)! \geq 2^n$, it suffices to choose $n = O(\log C)$. This gives an algorithm running in time $\mathit{poly}(\log C)$. Since the length of the encoding of $C$ is $\Theta(\log C)$, the resulting algorithm runs in polynomial time.