Inequality for Binomial Tail Probabilities Let $B(k,n,p)$ be the CDF of a binomial distribution. The following inequality appears to be true from numerical tests. 
$$B\left(k-1,n,\frac{k-1}{n-1}\right)>B\left(k,n,\frac{k}{n-1}\right)$$
I have looked through the relevant results summarized in section one of Xu, Balakrishnan (2011) to prove the result, but I have not found anything which can be applied easily. Guidance or helpful references are very welcome. 
Update:
I think the following may be helpful. I notice that $B\left(k-1,n,\frac{k-1}{n-1}\right)$ can be expressed as the probability of having no more than $k$ successes among $n+1$ Bernoulli variables where one has $p=1$ and the other have $p=\frac{k-1}{n-1}$.
$B\left(k,n,\frac{k}{n-1}\right)$ can be expressed as the probability of having no more than $k$ successes among $n+1$ Bernoulli variables where one has $p=0$ and the other have $p=\frac{k}{n-1}$.
 A: You could use the normal approximation to $N (\mu, \sigma^2) $ to get a solution that is valid for sufficiently large $n $. Reminding that the mean and variance for the binomial distribution are $np $ and $np (1-p) $, respectively,  in your inequality the first binomial distribution is approximated by
$$ N \left(\frac {n (k-1)}{n-1},  \frac {n(k-1)(n-k)}{(n-1)^2 } \right) $$ 
whereas the second one is approximated by 
$$ N \left(\frac {nk}{n-1},  \frac {nk \, (n-k-1)}{(n-1)^2} \right)$$
The areas of the CDF correspond to the probabilities given by the  $z$-values of the two normal distributions, calculated for $k-1$ and for $k $, respectively. So we obtain for the first distribution
$$\displaystyle \frac {(k-1)- \frac {n(k-1)}{n-1}  }{\sqrt{ \frac { n (k-1)(n-k) }{(n-1)^2}}}$$
$$\displaystyle =-\frac {\sqrt {k-1}}{\sqrt{ n (n-k)}} $$
and for the second one
$$\displaystyle \frac {k- \frac {nk}{n-1}  }{\sqrt{ \frac { nk(n-k-1) }{(n-1)^2}}}$$
$$\displaystyle =-\frac {\sqrt {k}}{\sqrt{ n (n-k-1)}} $$
Note that the two $z $-values are both negative. The initial inequality is then equivalent to
$$\displaystyle -\frac {\sqrt {k-1}}{\sqrt{ n (n-k)}} > -\frac {\sqrt {k}}{\sqrt{ n (n-k-1)}} $$
or
$$\displaystyle \frac {\sqrt {k-1}}{\sqrt{ n-k}} < \frac {\sqrt {k}}{\sqrt{ n-k-1}} $$
which is obviously true for any $n-k>1$ . 
It should be pointed out that using this method  the case $n-k=1$   is not defined because it leads to a division by zero. However, taking the right-handed limit we might consider that in this case the RHS tends to $\infty $, again satisfying the inequality. 
