<s>[INST] Show $x_n := \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1}$ is increasing in n for $p \in (0,1)$ I need to show that $x_n := \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1}$ is increasing in $n$ for $p \in (0,1)$. My attempts have involved trying to show
$\begin{eqnarray*} \frac{1-p^{n+2}}{1-p^{n+1}} \frac{n+1}{n+2} &>& \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1} \\
\left(1 + \frac{p^{n+1}}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2} &>& \left(1 + \frac{p^{n}}{\sum_{k=0}^{n-1} p^k} \right) \frac{n}{n+1} \\ 
\frac{n+1}{n+2} - \frac{n}{n+1} &>& \left(\frac{p^{n}}{\sum_{k=0}^{n-1} p^k} \right)\frac{n}{n+1}-\left(\frac{p^{n+1}}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2}\\ 
\frac{n+1}{n+2} - \frac{n}{n+1} &>& p^n \left[\left(\frac{1}{\sum_{k=0}^{n-1} p^k} \right)\frac{n}{n+1}-\left(\frac{p}{\sum_{k=0}^n p^k} \right) \frac{n+1}{n+2} \right]\end{eqnarray*}$
but this does not seem to simplify. 
I have also attmpted to treat $x_n$ as a continuous function in $n$ and differentiated (this leaves a mess so I won't put it up here). 
Sadly, I seem to be unable to get my required result from either. 
EDIT: It might be helpful to note that $x_n = \frac{\sum_{k=0}^n p^k}{\sum_{k=0}^{n-1} p^k}\frac{n}{n+1}$
EDIT 2: I can show that $x_n < 1$ for all $n \in \mathbb{N}$ and that $x_n \to 1$. But I don't think this suffices...
EDIT 3: If we can show that
$\frac{\sum_{k=0}^{n+1}p^k}{n+2} \cdot \frac{\sum_{k=0}^{n-1}p^k}{n} \geq  \left(\frac{\sum_{k=0}^{n+1}p^k}{n+2} \right)^2$, we are done. But I don't know how to show this.
 [/INST] Writing $x_n = \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}$, we see:
$$x_n>x_{n-1}\iff \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}>\frac{1-p^{n}}{1-p^{n-1}}\frac{n-1}{n}$$
$$\iff(1-p^{n-1})(1-p^{n+1})n^2>(1-p^n)^2(n^2-1)$$
$$\iff (1-p^n)^2 > n^2[(1-p^n)^2-(1-p^{n-1})(1-p^{n+1})]=n^2[p^{n-1}+p^{n+1}-2p^n]$$
$$\iff\frac{(1-p^n)^2}{p^n}>n^2\left(\frac{1}{p}+p-2\right)=n^2\frac{(p-1)^2}{p}$$
$$\iff \frac{1-p^n}{p^{n/2}}>n\frac{1-p}{p^{1/2}}\iff\frac{p^{n/2}}{p^{1/2}}\frac{1-p^{-n}}{1-p^{-1}}>n$$
$$\iff p^{\frac{n-1}{2}}(1+p^{-1}+p^{-2}+\cdots+p^{-(n-1)})>n$$
$$\iff \frac{1+p^{-1}+p^{-2}+\cdots+p^{-(n-1)}}{n}>p^{-\frac{n-1}{2}}$$
which is true by the AM-GM inequality.
</s>