Find asymptotic for $s(n)=\min\{m\in{\mathbb N}\mid C_n^m\cdot e^{-m^3/(\ln m)^{10}}<1\}$ I have some strange function: $s(n)=\min\{m\in {\mathbb N} \mid C_n^m\cdot e^{-m^3/(\ln m)^{10}}<1\}$ and I need to find asymptotics for it. I have a solution for this except one last step, I believe. So any help would be appreciated.
Solution is as follows: first we observe that $m = o(\sqrt n)$, in other cases ($m \ge \sqrt n$) exponent decreases much faster compared to growth of binomial coefficient. So, for the case of $m = o(\sqrt n)$ we could write following chain of (asymptotical) equalities:
\begin{align}
C_n^m\cdot e^{-m^3/(\ln m)^{10}} & \sim \frac{n^m}{m!} \cdot e^{-m^3/(\ln m)^{10}} \\                                 
    &= e^{m \ln n - \ln m! - m^3/(\ln m)^{10}} \\                                        
    &\sim e^{m \ln n - m \ln m \cdot (1 + o(1)) - m^3/(\ln m)^{10}} \\
\end{align}
For last equation to be less than $1$, exponent argument should be less than/asymptotically equal to $0$:
\begin{align}
& m \ln n - m \ln m \cdot (1 + o(1)) - \frac{m^3}{(\ln m)^{10}} \sim 0 \\                                             
& \ln n - \ln m \cdot (1 + o(1)) - \frac{m^2}{(\ln m)^{10}} \sim 0 \\                                                 
& \ln n - \frac{m^2}{(\ln m)^{10}} \cdot (1 + o(1)) \sim 0 \\
\end{align}
And in the last equation I should find $m$ in terms of $n$. And I don't know how can I do that. May be I missed something on the previous steps and $(\ln m)^{10}$ could be removed somehow. But I cannot see a way to do that.
Thanks in advance for any ideas.
 A: $\def\tfrac#1#2{{\textstyle\frac{#1}{#2}}}$
Your approach seems to be in the direct direction, but you need
to do more work and be more rigourous, especially when using asymptotic expansions of
binomial coefficients. You expanded $\log m!$ correctly keeping the asymptotic error term $o(1)$, but you didn't do the same for $n!/(n-m)!$, which isn't really correct.
Especially you need to keep trying different asymptotic forms for $m$; if $m=o(n^\gamma)$ for every $\gamma>0$, the next thing to try is always some power of $\log n$.
Also, it is wrong to write
$$ \text{anything} \sim 0, $$
because the definition of $f(x)\sim g(x)$ is $\lim_{x\to\infty}f(x)/g(x)=1$. The leading asymptotic term will be obtained from
$$ \text{something} \sim m^3/(\log m)^{10},$$
but even after the leading term is found, the exponent will not be asymptotically constant, because of smaller terms, so expecting the exponent to be $\sim 0$ is wrong.
First, because $n$ is large, we can ignore the way $m$ is
restricted to positive integers, and instead consider the equation
$$ {n\choose m}e^{-m^3/\log^{10} m} = 1 $$
with real $m$ and $n$. Taking the logarithm of it, we get
$$ \log\Gamma(n)-\log\Gamma(m)-\log\Gamma(n-m) = f(m), $$
where $f(m) = m^3/\log^{10}m$. Expanding using Stirling's
formula, and simplifying gives the following equation that must be satisfied by $m$:
$$ m\log(n/m) + O(m^2/n+m) = f(m). $$
Then proceed by trial and error. First, write $m\sim \rho n^\gamma$,
$0<\gamma<\frac12$:
$$ (1-\gamma)\rho n^\gamma\log n + O(n^\gamma+n^{2\gamma-1}) \sim n^{3\gamma}/\log^{10} n. $$
Clearly this cannot be solved (as you said), because it requires $\gamma=3\gamma$, $\gamma=0$, and we started with $\gamma>0$.
So, after some attempts, try the form $m\sim\rho (\log n)^\beta (\log\log n)^\delta$, which gives
$$ \rho (\log n)^{\beta+1}(\log\log n)^\delta \sim \frac{\rho^3 (\log n)^{3\beta} (\log\log n)^{3\delta}} { (\beta \log \log n)^{10} }, $$
which can be solved for the unknowns.
