Sum involving binomials and reciprocals

Question

Let $\rho \in (0,1)$ and $n \in \mathbb{N}$. I'm trying to show that $$ \frac{n(1+\rho)}{2^n} \sum_{i=0}^n \frac{\binom{n}{i}}{n+(3n-4i)\rho} \to 1 $$ as $n \to \infty$. I have no idea how to show this, I thought perhaps using the standard bounds for the binomial and upper bounding with an exponential would produce something helpful but I don't quite see hwo to make it work. Any ideas?

EDIT:

Using the binomial theorem I managed to write the expression as $$ \frac{n(1+\rho)}{2^n} \sum_{i=0}^n \frac{\binom{n}{i}}{n+(3n-4i)\rho} = \frac{n(1+\rho)}{2^n} \int_0^1 t^{3\rho n + n - 1}(1+t^{-4\rho})^n \, \mathrm{d}t $$

I thought this was perhaps helpful but the final integral evades me. Any ideas?

Answer 1

While the hint in @Greg Martin's comment is ok, it's a bit tedious. In fact, most of the heavy lifting is done already in the law of big numbers: if $X_i$ are i.i.d. random variables with $P(X_i=0)=P(X_i=1)=\frac12$ and $\displaystyle S_n=\sum^n_{i=1}X_i,$ we may recognize $$P(S_n=i)=2^{-n}\binom{n}{i},$$ so we have $$\frac{n(1+\rho)}{2^n} \sum_{i=0}^n \frac{\binom{n}{i}}{n+(3n-4i)\rho}=(1+\rho)\,\mathbb{E}\frac1{1+(3-4S_n/n)\rho}.$$

Now $S_n/n\to\frac12$ a.s. as $n\to\infty$ due to the law of large numbers, and from the dominated convergence theorem, we know that with $$Y_n=\frac1{1+(3-4S_n/n)\rho},$$ we have $$\lim_{n\to\infty}\mathbb{E}Y_n=\mathbb{E}\lim_{n\to\infty}Y_n,$$ since $1+(3-4S_n/n)\rho\ge1-\rho,$ i.e. $Y_n\le\frac1{1-\rho}.$ Thus, $$(1+\rho)\,\mathbb{E}\frac1{1+(3-4S_n/n)\rho}\to\frac{1+\rho}{1+(3-4\cdot\frac12)\rho}=1.$$