How does one find such a number? Not really one of those who usually ask for help here, but this case seems to be too much for me. I have been going over Courant’s “Differential and Integral Calculus”, and I have finally reached the problems section of the chapter 1.5 (i.e. “The limit of a sequence”). I would not have come here if it hasn’t been for the problem 9, namely the (e) part of it. The problem is generally about the sequence $a_n = \frac{10^n}{n!}$. I have, as all the parts (a)-(d) asked me to, found the limit of the mentioned sequence (=0), concluded whether it is monotonic or not, found the value of n such that the sequence is monotonic onwards and estimated the difference between the sequence and its limit respectively. Now, the (e) part demands that I calculate the exact value of n such that the difference mentioned is less than $\frac{1}{100}$. I have attempted to expand the factorial and try to deduce some helpful corollaries, but that does not seem to work. I am genuinely confused by this problem and not certain how I should approach it. It is of utmost importance that I note the following: I do not require the solution, I only need a HINT. Not a very crucial one, which virtually solves it (the problem), but one sufficient enough to proceed. I should be grateful for any help provided.
P.S. Please excuse me for some fairly probable mistakes in my writing (happens for I am not a native).
 A: Doing a binary search by hand shouldn't take long, and you don't need to include that work in your answer. Once you have $n$ such that $\frac{10^{n-1}}{(n-1)!}>\frac1{100}>\frac{10^n}{n!}$, you have the answer with proof.
A: I'll answer for the inverse ratio, $\dfrac{n!}{10^n}$.
Taking the logarithm and using the simplest Stirling approximation, we first solve
$$n\ln n-n-n\ln10\approx0,$$ or $$n\approx 100e=27.18\cdots$$ which tells us where the ratio is near $1$ (and we still need a factor $100$). So $n=27$ is a by-default approximation.
As more precisely $$\frac{27!}{10^{27}}=10.88\cdots$$ it suffices to evaluate the partial products of
$$\frac{27!}{10^{27}}\cdot2.8\cdot2.9\cdot3.0\cdot3.1\cdots$$ and three factors will do.
A: How to narrow it down. A crude lower bound: If $n\ge 2$ then  $$\ln (n!)=\sum_{j=2}^n\ln j>$$ $$>\sum_{j=2}^n\int_{j-1}^j(\ln t)dt=$$ $$=\int_1^n(\ln t)dt=(n\ln n)-n+1=$$ $$=F(n).$$
We have $a_n<1/100\iff (n+2)\ln 10<\ln (n!).$ Hence  $(n+2)\ln 10<F(n)\implies a_n<1/100.$ And we have $$(n+2)\ln 10<F(n)\iff (1+\frac2n) \ln (10e) <(\ln n)+\frac 3n.$$ Since $\ln (10e)\approx 3.3$ and $\ln (36)=2\ln 2+2\ln 3\approx (2)(0.7)+2(1.1)=3.6,$ it appears that $n\ge 36$ would suffice to make $a_n<1/100$.  But $F(n)$ is a crude lower bound for $\ln (n!)$ so it seems we should work $down$ from $36.$
In fact we have $F(30)<32\ln 10<\ln (30!).$
