Find the constant $c$ in the equation $$\max_{a\le n/2}\frac{C_n^a}{\sum_{k=0}^{\lfloor{a/3}\rfloor}C_n^k}=(c+o(1))^n.$$
I've tried to use this asymptotics $$C_n^k \sim \frac{n^m}{m!} \sim e^{m\ln n - \ln m!} \sim e^{m \ln n - m\ln m (1 + o(1))}$$
I've got: $$\max_{a\le n/2}\frac{C_n^a}{\sum_{k=0}^{\lfloor{a/3}\rfloor}C_n^k}=\max_{a\le n/2}\frac{e^{a\ln n - a \ln a(1+o(1))}}{\sum_{k=0}^{\lfloor{a/3}\rfloor}e^{k\ln n - k\ln k(1+o(1))}} = \max_{a\le n/2} e^{a\ln n - a \ln a(1+o(1)) - \sum_{k=0}^{\lfloor{a/3}\rfloor}(k\ln n - k\ln k(1+o(1)))} $$
Now I need to find maximum of function $$f(n, a) = a\ln n - a \ln a(1+o(1)) - \sum_{k=0}^{\lfloor{a/3}\rfloor}(k\ln n - k\ln k(1+o(1))) , a \le \frac{n}{2}$$
Please, help me to analyse this function. Or may be I should use another asymptotcs to simplify this equaition?
Since this is a homework, I'll give a sketch of the proof.
Since $C_n^p\leq\sum\limits_{k=0}^{p} C_n^k\leq (p+1) C_n^p$ for any $0\leq p\leq n$, then $$ \sum\limits_{k=0}^{p} C_n^k=C_n^p f(p) $$ for some function $f$ such that $1\leq f(p)\leq 1+p$. Since $f=(1+o(1))^n$ and we want to get the final asymptotics of the form $(c+o(1))^n$, then it is out of our interest. Thus we need to find asymptotics of $$ \max_{0\leq a\leq n/2} \frac{C_n^a}{C_n^{a/3}}=\max_{0\leq\lambda\leq1/2}\frac{C_n^{[n\lambda]}}{C_n^{[n\lambda/3]}} $$ As you already should know $$ C_n^{[n\lambda]}=(e^{H(\lambda)}+o(1))^n $$ for $H(\lambda)=-\lambda\ln (\lambda)-(1-\lambda)\ln(1-\lambda)$, then it is enough to maximize on $[0,1/2]$ the function $H(\lambda)-H(\lambda/3)$. The maximum is attained for $\lambda=1/3$ and equals to $\ln(4/3)$. So the desired asymptotics is $$ \max_{0\leq a\leq n/2}\frac{C_n^a}{\sum\nolimits_{k=0}^{[a/3]}C_n^k} =(e^{\ln(4/3)}+o(1))^n =(4/3+o(1))^n $$