Here's one from an old paper of mine. It has the property of being precise all the way from the middle to the end.Define$$ Y(x) = e^{x^2/2}\int_x^\infty e^{-t^2/2}dt. $$Define $x=(2k-n)/\sqrt{n}$. Then for $x\ge 0$,$$\sum_{j=k}^n \binom nj = \sqrt{n} \,\binom{n-1}{k-1} Y(x)\, e^{E(k,n)/\sqrt n},$$where $0\le E(k,n)\le\min(\sqrt{\pi/2},2/x)$.If $x<0$, use $\sum_{j=k}^n \binom nj=2^n-\sum_{j=n-k+1}^n \binom nj$.
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