I'm looking for a tight bound on: $$\sum_{k=0}^r {n\choose k}$$ where I can assume that $r<n$.
Looking at the pascal triangle I understand that $n\choose n/2$ might be the largest element among the row, hence $r {n\choose n/2}$ might work. This expression can further be simplified with Stirling approximation.
However, is there a tighter bound?