<s>[INST] Finding $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$ Find $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$.
I wrote this proof:
Let $f(x)=e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
$|\sqrt{e}-P_N(\frac{1}{2})|=|R_N(\frac{1}{2})| < \frac{1}{100}$
From the Taylor theorem we get that there exists $0<c<\frac{1}{2}$ such that:
$|R_N(\frac{1}{2})| = |\frac{f^{(N+1)}(c)}{(N+1)!2^N}| = \frac{e^c}{(N+1)!2^N} \leq \frac{e}{(N+1)!2^N}\leq \frac{3}{(N+1)!2^N}\leq \frac{3}{(N+1)!}<10^{-2}$
hence for $N=5$ we get $720>300$ and the inequality holds.
Therefore:
$P_5(\frac{1}{2})=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{6\cdot8}+\frac{1}{16\cdot 24}+\frac{1}{32\cdot 120} = \frac{32\cdot120+16\cdot 120+4\cdot 120+4\cdot 20+2\cdot 5}{32\cdot 120} = \frac{n}{m}$
Did I get it right? Was there a simpler method?
 [/INST] Your solution is correct, but unnecessarily complicated. Also, there seems to be an error in your expression for $\left\lvert{R_N\left(\tfrac12\right)}\right\rvert,$ which doesn't invalidate your numerical answer, but adds to the complication of finding it.
You're using Taylor's theorem with the Lagrange form of the remainder. Surely the expression should be:
$$
\left\lvert{R_N\left(\frac12\right)}\right\rvert = \left\lvert{\frac{f^{(N+1)}(c)}{(N+1)!\cdot2^{N+1}}}\right\rvert
$$
Having $2^{N+1}$ rather than $2^N$ in the denominator makes things slightly easier. Also, you have gone rather too far in your simplifying approximations; the approach is valid, but the calculation is harder than it needs to be.
Continuing in a simpler way, using the fact that $e < 4,$ so $\sqrt{e} < 2$:
$$
\frac{e^c}{(N+1)!\cdot2^{N+1}} < \frac{\sqrt{e}}{(N+1)!\cdot2^{N+1}} < \frac{1}{(N+1)!\cdot2^N}.
$$
Therefore, all we need is to find $N$ large enough that:
$$
(N+1)!\cdot2^N > 100.
$$
So we don't need $N = 5$; it is good enough to take $N = 3.$ Then we find:
$$
\sqrt{e} - \frac{79}{48} < \frac{1}{192}.
$$
We could get the same value for $\tfrac{n}{m}$ even more simply, without using the remainder expression for Taylor's theorem - it's quite easy to get it wrong, and I'm not sure I haven't! - instead just using the infinite series for $\sqrt{e}$:
\begin{align*}
\sqrt{e} & = 1 + \frac12 + \frac18 + \frac1{48} + \frac1{384}\left(1 + \frac1{5\cdot2} + \frac1{5\cdot6\cdot2^2} + \cdots\right) \\
& < \frac{79}{48} + \frac1{384}\left(1 + \frac1{2} + \frac1{2^2} + \cdots\right) \\
& = \frac{79}{48} + \frac1{192},
\end{align*}
an approximation which is still more than good enough $\ldots$ indeed, the accuracy is just the same! (I miscalculated at first.) I presume this is just a coincidence.
(Continued fraction methods, as used in the two previous answers, give more accurate approximations than these - in fact, the most accurate approximations possible - but we don't need that level of precision for this question.)
</s>