I am studying the function $U: \mathbb{R}_{\ge 0} \to \mathbb{R}$ given by \begin{align} U(l) &= (1-r^l) \left[ T - \tau n^l \left( 1 - \left( 1 - \frac{1}{\tau n^l} \right)^T \right) \right] \\ &=(1-r^l)\sum_{i=2}^T \binom{T}{i}(-1)^i\frac{1}{(\tau n^l)^{i-1}} \end{align} where $0 < r < 1$ and $\tau \in\mathbb{Z}_{\ge 1}$ and $n, T \in \mathbb{Z}_{\ge 2}$ are fixed, with $T>\tau$. This is a utility function in an application.
It seems that, regardless of the values of the parameters $r,l,T$, and $\tau$ (within their respective bounds), $U(l)$ starts at zero (for $l=0$), and for some $l_0>0$ increases in $[0,l_0]$, has a unique local (and global) maximum at $l=l_0$ and is decreasing in $[l_0,\infty]$ and tends to zero as $l\to\infty$. (See example graph in link below.)
I want to prove the part that $U'(l)=0$ has exactly one root (at some $l_0>0$).
I have not gotten anywhere with the equation $U'(l)=0$ as is. Neither with $U'(l_0)=U'(l_1)=0$ for $l_0 \ne l_1$ leading to a contradiction. For $T=2$ it can be solved analytically. Yet another idea is therefore a proof by induction over $T$, but I cannot see how $U'(l)|_{T=t}$ having exactly one zero leads to $U'(l)|_{T=t+1}$ having exactly one zero.
However, I have been able to prove that $U'(0)>0$ and that there exists an $L>0$ such that $U'(l)<0$ for all $l\ge L$. By the intermediate value theorem, $U'$ (which is continuous) must then have a zero in $(0,L)$. The question is how to prove that it is only one. If one could establish $U'(l_0)=0 \Rightarrow U''(l_0)<0$ we would be done, since $U'$ is continuous, but I cannot see how to use $U'(l_0)=0$ to prove $U''(l_0)<0$.
First try: The variable substitution $\tau n^l=x$ induces the new function $u: \mathbb{R}_{\ge\tau} \to \mathbb{R}$ given by \begin{align} u(x) &= (1-\tau^B x^{-B}) \left[ T - x \left( 1 - \left( 1 - \frac{1}{x} \right)^T \right) \right] \\ &= \sum_{i=2}^T(-1)^i\binom{T}{i} x^{-i+1} - \sum_{i=2}^T(-1)^i\binom{T}{i} \tau^B x^{-i+1-B} \end{align} (where $B = -\frac{\log r}{\log n} > 0$) whose derivative $u'(x)$ (after a binomial expansion) can be written $$ u'(x)=p(x)+q(x) $$ where $p(x)=\sum_{i=2}^T p_i x^{-i}$ and $q(x)=\sum_{i=2}^T q_i x^{-i-B}$ are generalized polynomials with $$ p_i=(-1)^i (-i+1) \binom{T}{i} \quad \text{and} \quad q_i=-(-1)^i (-i+1-B) \binom{T}{i} \tau^B. $$ For example, if $\tau = 2, n = 10, r = 0.7$ and $T = 5$, then (with rounded off values) $B=0.15$ and $$ u'(x) = -10x^{-2} + 12.86x^{-2.15} + 20x^{-3} - 23.99x^{-3.15} - 15x^{-4} + 17.56x^{-4.15} + 4x^{-5} - 4.63x^{-5.15}. $$
Since $u'(x)$ is a generalized polynomial, I thought first of using [Graham J.O. Jameson, Counting zeros of generalised polynomials: Descartes’ rule of signs and Laguerre’s extensions, The Mathematical Gazette 90 (2006), no. 518, 223–234]. Unfortunately, neither the generalized Decartes' rule of signs (Theorem 3.1) nor the theorem about sign changes in partial coefficient sums for zeros in $(1,\infty)$ (Theorem 4.7) is of any use as $u'(x)$ in general has more than one (cumulative) sign change.
Second try: Use the same substitution as above and write the equation $u'(x)=0$ as $g(x)=x$ and check if $g:\mathbb{R}_{\ge\tau}\to\mathbb{R}_{\ge\tau}$ is a contraction (i.e. there is some real number $0\le k<1$ such that $|g(x)-g(y)|<k|x-y|$ for all $x,y\in\mathbb{R}_{\ge\tau}$). If so, it would follow from Banach's fixed point theorem that $u'(x)=0$ has exactly one root. Unfortunately $g'(\tau)$ is not $<1$ in general, so $g$ is not a contraction.
Are there any other ways or perhaps different approaches altogether to prove that $U'(l)$ (or $u'(x)$) has exactly one zero?
The second challenge is to prove that $l_0$ increases with $T$ (while keeping all the other parameters fixed). From simulations, this seems to be the case, but I currently have no idea how to prove it.