How to show this specific sequence of functions is increasing. In studying for an exam, I've come across a fact that is clear and yet frustrating difficult to prove. I need to show that 
$\chi_{[0,k]}(x)\left(1-\frac{x}{k}\right)^{k}\le \chi_{[0,k+1]}(x)\left(1-\frac{x}{k+1}\right)^{k+1}$
for every $k\in \mathbb{N}$ (here $\chi$ denotes a characteristic function). I'd appreciate a hint or a solution, but so far expanding everything out has been a nightmare, and I'm afraid that I'm missing a simple solution. 
Thanks.
 A: The equation obviously holds for $x < 0$ and $x > k$, so what remains is the case $0 \leq x \leq k$. Since $\chi_{[0,k]}(x)=\chi_{[0,k+1]}(x) = 1$ on this interval, what remains is to show that
$$\left( 1 - \frac{x}{k} \right)^k \leq \left( 1 - \frac{x}{k+1} \right)^{k+1}$$
for $0 \leq x \leq k$.
Now fix $x$ and consider the function $f(k) = \left( 1- \frac{x}{k} \right)^k$. If we can show that it is increasing, then we're done. In this case it's better to show that $\ln f(k)$ is increasing, which is equivalent. Have you tried this?
A: For the sake of notation, call the members of your sequence $f_k$. Since $f_k$ dies on $[k,k+1]$ and every function is non-negative, it suffices to show that $1\le \frac{f_{k+1}}{f_k}$ when $0\le x\le k$. So let's play some worst case-scenario$^*$ to get the bound. 
\begin{align*}\frac{f_{k+1}}{f_k}&=(1-\frac{x}{k+1})\left(\frac{1-\frac{x}{k+1}}{1-\frac{x}{k}}\right)^k\\ &=(1-\frac{x}{k+1})(\frac{k}{k+1})(1+\frac{1}{k-x})\\ &\overset{*}{\ge} (1-\frac{0}{k+1})(\frac{k}{k+1})(1+\frac{1}{k-0})\\ &=(\frac{k}{k+1})(1+\frac{1}{k})\\ &=1,\end{align*}
as desired. 
