Let $f_p(k) = -\frac{(1-p)^k}{k \ln(p)}.$ Show that $0 \le f_p(k) \le 1$. For each $p \in (0,1)$, define the function (or, if you prefer, sequence) $f_p:\mathbb{N} \to \mathbb{R}$ by
$$f_p(k) = -\frac{(1-p)^k}{k \ln(p)}. $$
I'm trying to show that, for any $p \in (0,1)$, we have $0 \le f_p(k) \le 1$, for every $k \in \mathbb{N}$.
Let $p \in (0,1)$ be given. Then $\ln(p) < 0$; and so, it is not too hard to see that $f_p(k) \ge 0$ for every $k \in \mathbb{N}$. However, I am having a hell of a time trying to show that $f_p(k) \le 1$.
My plan was to just show that $f_p(1) \le 1$, and then show that $f_p$ is decreasing (by showing that $f'_p$ is negative). Both of these facts are quite evident from the graph of $f_p$; but proving them has been rather difficult, for whatever reason...
 A: The function, or sequence, you have is
$$f_p(k) = -\frac{(1-p)^k}{k\ln(p)} \tag{1}\label{eq1A}$$
This gives
$$f_p(1) = \frac{1-p}{-\ln(p)} \tag{2}\label{eq2A}$$
Using a definition of the natural logarithm gives, since $p \in (0, 1)$, that
$$\ln(p) = \int_{1}^{p}\frac{dx}{x} \implies -\ln(p) = \int_{p}^{1}\frac{dx}{x} \tag{3}\label{eq3A}$$
With $0 \lt p \le x \lt 1 \implies \frac{1}{x} \gt \frac{1}{1} = 1$ within the range being integrated gives the inequality
$$\int_{p}^{1}\frac{dx}{x} \gt \int_{p}^{1}dx = 1 - p \tag{4}\label{eq4A}$$
Combined with \eqref{eq3A}, along with that the quantities are positive, gives
$$-\ln(p) \gt 1 - p \implies \frac{1}{-\ln(p)} \lt \frac{1}{1 - p} \tag{5}\label{eq5A}$$
Substituting into \eqref{eq2A} gives
$$f_p(1) \lt \frac{1 - p}{1 - p} = 1 \tag{6}\label{eq6A}$$
Next, using that $1 - p = e^{\ln(1 - p)}$ in \eqref{eq1A}, along with other manipulations, gives
$$f_p(k) = -\frac{1}{\ln(p)}\left(e^{\ln(1-p)k}\right)(k^{-1}) \tag{7}\label{eq7A}$$
Differentiating \eqref{eq7A} using the product & chain rules gives
$$\begin{equation}\begin{aligned}
f'_p(k) & = -\frac{1}{\ln(p)}\left(\ln(1-p)\left(e^{\ln(1-p)k}\right)(k^{-1}) - \left(e^{\ln(1-p)k}\right)k^{-2}\right) \\
& = -\frac{e^{\ln(1-p)k}}{k\ln(p)}\left(\ln(1-p) - \frac{1}{k}\right)
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
Since $\ln(p) \lt 0$, the outside factor $-\frac{e^{\ln(1-p)k}}{k\ln(p)} \gt 0$. Inside the brackets, $\ln(1 - p) \lt 0$, so the result there is negative, giving that overall
$$f'_p(k) \lt 0 \tag{9}\label{eq9A}$$
i.e., $f_p(k)$ is strictly decreasing. As you pointed out, $f_p(k) \ge 0$ for all $k \in \mathbb{N}$, with it actually being $f_p(k) \gt 0$. Thus, along with \eqref{eq6A} and \eqref{eq9A}, gives
$$0 \lt f_p(k) \lt 1 \tag{10}\label{eq10A}$$
Update: Another way to show the derivative is always negative it to first take the natural logarithm of both sides of \eqref{eq1A} to get
$$\ln(f_p(k)) = k\ln(1-p) - \ln(k) - \ln(-\ln(p)) \tag{11}\label{eq11A}$$
Next, differentiating wrt $k$ gives
$$\frac{f'_p(k)}{f_p(k)} = \ln(1-p) - \frac{1}{k} \tag{12}\label{eq12A}$$
Using $\ln(1-p) \lt 0$ and $f_p(k) \gt 0$ gives \eqref{eq9A}.
