I have the following function: $ f(p) = \sum^{k-1}_{i=0} \binom{k-1}{i} p^{i} (1-p)^{k-i-1} (i+1)^{-1} $
by going through random values of k and $p \in [0,1]$, i've seen that as p increases, $f(p)$ is decreasing. Therefore, I wanted to prove that. The second derivative of $f(p)$ in respect to $p$ was a good place to start, but the final equation $\frac{d^2 f(p)}{dp^2}$ is too cumbersome and messy to work with and I ended up with nothing.
Can someone guide me to the right way of proving this? :)
First simplify the summation:
$$\begin{align*} \sum_{i=0}^{k-1}\frac1{i+1}\binom{k-1}ip^i(1-p)^{k-1-i}&=\frac1k\sum_{i=0}^{k-1}\binom{k}{i+1}p^i(1-p)^{k-1-i}\\ &=\frac1k\sum_{i=1}^k\binom{k}ip^{i-1}(1-p)^{k-i}\\ &=\frac1{kp}\sum_{i=1}^k\binom{k}ip^i(1-p)^{k-i}\\ &=\frac1{kp}\left(1-(1-p)^k\right)\;. \end{align*}$$
The first step uses the identity $\binom{k}{i+1}=\frac{k}{i+1}\binom{k-1}i$, and the last is the binomial theorem. Now take the derivative with respect to $p$ to get
$$-\frac1{kp^2}+\frac{kp(1-p)^{k-1}+(1-p)^k}{kp^2}\;;$$
you’d like to show that this is less than or equal to $0$. After a little algebra that reduces to showing that
$$(1-p)^{k-1}\big((k-1)p+1\big)\le 1\;.$$
But $1+(k-1)p\le(1+p)^{k-1}$, so
$$(1-p)^{k-1}\big((k-1)p+1\big)\le(1-p)^{k-1}(1+p)^{k-1}=\left(1-p^2\right)^{k-1}\le 1\;,$$
as desired.