Estimating the limit $\lim_{n\to\infty}\frac{n _2F_1[1 - n, 1 + 2^n n; 2 + 2^n n; -1]}{1 + 2^n n}$

Question

Can you please help me solve the limit below? $$\lim_{n\to\infty}\frac{n _2F_1[1 - n, 1 + 2^n n; 2 + 2^n n; -1]}{1 + 2^n n}$$ where $_2F_1(a,b;c;z)$ - hypergeometric function

Answer 1

The hypergeometric function is a polynomial: \begin{eqnarray*} & & {}_2F_1[1 - n, 1 + 2^n n; 2 + 2^n n;z]\\ &=&\sum_{k\ge 0} \frac{(1-n)_k (1+2^n n)_k}{(2+2^n n)_k k!} z^k\\ & =& \sum_{0\le k\le n-1} \frac{(-1)^k(n-1)\cdots (n-k) (2^n n+1)}{(2^n n + k + 1) k!} z^k\\ &=& \sum_{0\le k\le n-1} \binom{n-1}{k} \left(1-\frac{k}{2^n n + k + 1}\right)(-z)^k \end{eqnarray*} so \begin{eqnarray*} && n {}_2F_1[1 - n, 1 + 2^n n; 2 + 2^n n;-1]\\ &=& n \sum_{0\le k\le n-1} \binom{n-1}{k} \left(1-\frac{k}{2^n n + k + 1}\right)\\ &=& 2^{n-1}n - n \sum_{0\le k\le n-1} \frac{k}{2^n n + k + 1}\binom{n-1}{k}\\ &=& 2^{n-1}n + O(n), \end{eqnarray*} meaning that the limit is $\frac 12 $.