# Series limit involving Binomial coefficients

Consider the parameters $$v_{a}$$, $$v_{b}$$ be such that $$0 and $$c>0$$. I have an equation involving the Binomial distribution that I need to solve with respect to $$p(n)$$:

$$\sum_{k=1}^{n}\left(\begin{array}{c} n-1\\ k-1 \end{array}\right)p^{k-1}\left(1-p\right)^{n-k}\cdot\left\{ v_{a}\left(\frac{1}{k}-\frac{k-1}{c\,k^{2}}\right)\right\} =\sum_{k=1}^{n}\left(\begin{array}{c} n-1\\ k-1 \end{array}\right)\left(1-p\right)^{k-1}p^{n-k}\cdot\left\{ v_{b}\left(\frac{1}{k}-\frac{k-1}{c\,k^{2}}\right)\right\}$$

I can show that this solution has a unique solution $$p\left(n\right)\in\left(0,1\right)$$. (Assume that $$c$$ is such that each terms in brackets are positive for all $$k$$.) I wish to understand the solution to this equation, $$p\left(n\right)$$, in the limit $$n\rightarrow\infty$$.

I conjecture that $$\lim_{n\rightarrow\infty}p\left(n\right)=\frac{1}{2}$$. Any way to show this? Hint: The terms in brackets strictly decrease in $$k$$ and as $$n$$ grows the Binomial coefficients shift to the right'', so each of these sums should go to zero. Is this enough? I am concerned with the rate they go to zero... Thank you in advance for any inputs.

The answer to the question is that $$\lim_{n\rightarrow\infty}p\left(n\right)=\frac{v}{v+1}$$, where $$v=\frac{v_{a}}{v_{b}}$$. That is, my conjecture is correct, if and only if, $$v_{a}=v_{b}$$. (Existence and uniqueness) $$p\left(\cdot\right)$$ is bounded and strictly increasing. Hence, it converges to its supremum, which is unique. To explicitly find the limit $$\lim_{n\rightarrow\infty}p\left(n\right)$$, one has to use the fact that the LHS is the expectation of indenpendent non-IID $$\text{Bin}\left(n,p\left(n\right)\right)$$ and check that a suitable version of the CLT applies.