Derive maximal sum of binomial terms How to determine a positive value of variable $m$, so that the following formula is maximized.
$$\frac{(1-q)^m}{\sum_{x=m/c}^{m}{\binom{m}{x} (1-q)^{m-x} q^x}}$$
where $1<c<m$, $0<q<1$, both are constants.
 A: Le $R_m$ denote the ratio considered, $R^*=\sup_mR_m$ and $r=(1-q)/q$.
If $r\le1$, $R_m\le1$ for every $m\ge0$ and $R_0=1$ hence $R^*=1$. Note that $R_1=r$ hence $R^*>1$ for any $r>1$.
If $r>2^c$, $R_m\to+\infty$ when $m\to+\infty$ hence $R^*=+\infty$.
Using large deviations estimates, one can refine this last result as follows: for any $c>1$, if $r>\varrho(c)$, then $R_m\ge k(r)^{m/c}$ where $k(r)>1$, hence $R^*=+\infty$, with
$$
\varrho(c)=c^c/(c-1)^{c-1}.
$$
Likewise, if $r<\varrho(c)$, then $R_m=k(r)^{(m/c)+o(m)}$ where $k(r)<1$, hence $R^*$ is finite.
One can guess that, for every $m\ge0$, $R^*=R_m$ for any $r$ in $(\varrho_{m}(c),\varrho_{m+1}(c))$, where $(\varrho_m(c))_{m\ge0}$ is an increasing sequence such that $\varrho_0(c)=0$, $\varrho_1(c)=1$ and $\varrho_m(c)\to\varrho(c)$ when $m\to+\infty$.

Edit Here is a sketch of the large deviations estimate. Let $X$ and $(X_k)$ denote i.i.d. Bernoulli random variables with $P(X=1)=q$ and $P(X=0)=1-q$, and $S_m=X_1+\cdots+X_m$. Then 
$$
R_m=(1-q)^m/D_m,\qquad D_m=P(S_m\ge m/c).
$$
By the law of large numbers, $D_m\to1$ if $q>1/c$ and $D_m\to0$ if $q<1/c$. From now on, we assume that $q<1/c$. Then $D_m$ is geometricaly small in the sense that $D_m\le a(c)^{m/c}$ where $a(c)<1$ can be determined by a standard application of Cramér inequality, as follows. 
Let $u>1$. Then $[S_m\ge m/c]=[u ^{S_m}\ge u^{m/c}]=[u ^{S_m-m/c}\ge 1]$ hence
$$
D_m\le u^{-m/c}E(u^{S_m})=\left(u^{-1}E(u^{X})^c\right)^{m/c}.
$$
This holds for every $u>1$ hence  $D_m\le a(c)^{m/c}$ with
$$
a(c)=\inf_{u>1}[u^{-1}E(u^{X})^c]=\inf_{u>1}[u^{-1}(qu+1-q)^c].
$$
The infimum is achieved at $u(c)$ which solves the equation $q(c-1)u=1-q$, and $u(c)>1$ for every $1<c<1/q$, hence
$$
a(c)=u(c)^{-1}(qu(c)+1-q)^c.
$$
This shows that $R_m\ge k(r)^{m/c}$ with $k(r)=(1-q)^c/a(c)$. Some easy computations yield
$$
k(r)=((c-1)/c)^c(r/(c-1))=r/\varrho(c).
$$
This proves that $R_m\to+\infty$ as soon as $k(r)>1$, that is, as soon as $r>\varrho(c)$.

Second edit Each $\varrho_m(c)$ is the point where $R_m$ replaces $R_{m-1}$ as the maximum. In particular, at $r=\varrho_m(c)$, $R_m=R_{m-1}$. This is equivalent to $\varphi_m(r)=\varphi_{m-1}(r)$, where
$$
\varphi_m(r)=\sum_{x=m/c}^m{m\choose x}r^{-x}.
$$
For example, for every positive $r$, $\varphi_0(r)=1$ and $\varphi_1(r)=1/r$. One sees that $\varphi_1(r)>\varphi_0(r)$ until $r=1$ where $\varphi_1(r)=\varphi_0(r)$, hence $\varrho_1(c)=1$.
