Inequality with Sum of Binomial Coefficients

It looks that the same was proven in the paper E.L.Johnson, D.Newman, K.Winston. An Inequality on Binomial Coefficients. Annals of Discrete Mathematics 2 (1978) 155-159. https://doi.org/10.1016/S0167-5060(08)70330-3 page 1 page 2 page 3 page 4 page 5

This is partially a comment that is slightly too long. In the Math Overflow article, we want to bound

$$ {{N \choose k} + {N \choose k-1} + {N \choose k-2}+\dots \over {N \choose k}} = {1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots} $$

The author chooses to use a geometric series as an upper bound starting with the first term. However, we can slightly delay the geometric series to get a smaller upper bound as follows:

$$ 1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots = \\1 + \frac{k}{N-k+1} \left(1 + \frac{k-1}{N-k+2} + \frac{(k-1)(k-2)}{N-k+2(N-k+3)} \right) + \cdots$$ Using a geometric series for an upper bound in the inner parenthesis gives us an upper bound of $$ \dbinom{N}{k} \left(1 + \frac{k}{N-k+1} \cdot \frac{N-k+2}{N-2k+3} \right). $$ We can easily check that for large enough $N$, $$1 + \frac{k}{N-k+1} \cdot \frac{N-k+2}{N-2k+3} < \frac{N-(k-1)}{N-(2k-1)}. $$ Finally we can check that the steps used by the OP in the 2nd attempt portion of the question does go through if $N = 3k$ (which was an issue last time) for large enough $N$. This is just some algebraic manipulations so I won't post it here. Hopefully I haven't made an error.