Monotonicity of a fraction combined with series (related to probability distributions) Let $(p_n)_{n \geq 0}$ be a probability distribution on $\mathbb{N}_0$ with finite expectation, thus $\sum_{n = 0}^\infty p_n = 1$ and $\sum_{n=0}^\infty n \, p_n < \infty$. I want to show that for all $0 \leq s \leq t < 1$ $$ \frac{\sum_{n=1}^{\infty} (1-s^n)\,  n \, p_n}{\sum_{n=1}^{\infty} (1-s^n)\, p_n} \leq \frac{\sum_{n=1}^{\infty} (1-t^n)\,  n \, p_n}{\sum_{n=1}^{\infty} (1-t^n)\, p_n}. $$
It is actually a little bit surprising to me that this is true as I thought that the $n$ in the numerator increases the weight of the bigger choice of $t$ compared to the denominator, but I plotted the function for some choices of $(p_n)$ and found that it is indeed increasing. 
Any ideas for formal proofs or arguments are appreciated.
 A: Since you seem surprised by the result, here is a non-rigorous, but maybe intuitively helpful, explanation.
This argument uses three independent r.v.s, $X, Y, Z$.  Let $X$ be chosen according to $p_n$, i.e. $P(X = n) = p_n$.
Let $Y$ be a geometric r.v. as follows.  Keep flipping a biased coin with $P(Head) = s$ until you see the first Tail.  Let $Y =$ no. of Heads before that first Tail.  We have $P(Y \ge n) = s^n$ and $P(Y < n) = 1 - s^n$.
Combining, we have $P(Y < X = n) = (1 - s^n) p_n$, which is the term appearing in your summation.  In particular the LHS denominator $=\sum (1-s^n) p_n = P(Y < X)$.  So the whole LHS can be interpreted as
$$LHS = E[X \mid X > Y]$$
Let $Z$ be a geometric r.v. defined similarly to $Y$, except this second coin has $P(Head) = t$, then the RHS is $E[X \mid X > Z]$.
Now if $t > s$, then you are likely to see more Heads from the second coin than from the first.  So intuitively speaking, conditioning on $X > Z$ will "push" $X$ to bigger values than conditioning on $X > Y$, i.e.
$$RHS = E[X \mid X > Z] \ge E[X \mid X > Y] = LHS$$
Let me stress again that this is not a proof, but rather, just an intuitive explanation.  Hope it helps anyway!
A: Note that for $m > n\ge 1$, $$
\frac{1-x^m}{1-x^n}
$$
is an increasing function of $x$ for $x\in(0,1)$ (see bottom for proof). Hence all $m>n$ and $t \ge s$:
\begin{eqnarray}
(m-n)\frac{1-s^m}{1-s^n} &\le& (m-n)\frac{1-t^m}{1-t^n}\\
(m-n)(1-s^m)(1-t^n) &\le& (m-n)(1-t^m)(1-s^n)\\
n (1-s^n)(1-t^m) + m(1-s^m)(1-t^n) &\le& m(1-t^m)(1-s^n) + n(1-t^n)(1-s^m)
\end{eqnarray}
Multiplying through by $p_np_m$, we obtain:\begin{eqnarray}
&&p_np_mn (1-s^n)(1-t^m) + p_np_m m(1-s^m)(1-t^n) \\&\le& p_np_mm(1-t^m)(1-s^n) + p_np_mn(1-t^n)(1-s^m)
\end{eqnarray}
This is true for all $m>n\ge 1$, so we can sum over all such pairs to obtain \begin{eqnarray}
&&\sum_{m>n\ge 1} \left(p_np_mn (1-s^n)(1-t^m) + p_np_m m(1-s^m)(1-t^n)\right)\\ &\le& \sum_{m>n\ge 1}\left( p_np_mm(1-t^m)(1-s^n) + p_np_mn(1-t^n)(1-s^m)\right)
\end{eqnarray}
Manipulating the left-hand side: Since the sum converges absolutely, we can split the term $p_np_mn (1-s^n)(1-t^m) + p_np_m m(1-s^m)(1-t^n)$ into separate terms for $(n,m)$ and $(m,n)$, yielding 
\begin{eqnarray}
&&\sum_{m>n\ge 1} \left(p_np_mn (1-s^n)(1-t^m) + p_np_m m(1-s^m)(1-t^n)\right)\\
&=&\sum_{m>n\ge 1} p_np_mn (1-s^n)(1-t^m) +\sum_{m>n\ge 1} p_np_m m (1-s^n)(1-t^m)\\ &=& \sum_{m>n\ge 1} p_np_mn (1-s^n)(1-t^m) +\sum_{n>m\ge 1} p_mp_n n (1-s^m)(1-t^n)\\
&=&\sum_{n,m\in \mathbb{N}, n\ne m} p_np_mn (1-s^n)(1-t^m)
\end{eqnarray}
We can do the same to the right hand side to obtain the inequality:$$
\sum_{n,m\in \mathbb{N}, n\ne m} p_np_mn (1-s^n)(1-t^m) \le \sum_{n,m\in \mathbb{N}, n\ne m} p_np_m m(1-s^n)(1-t^m)
$$
Since the excluded diagonal terms (i.e. the ones with $n=m$) would be equal in both sums, we can add them to both sides and get the inequality:$$
\sum_{n,m=1}^\infty p_np_mn (1-s^n)(1-t^m) \le \sum_{n,m=1}^\infty p_np_m m(1-s^n)(1-t^m)
$$
Which allows us to complete the proof with some elementary manipulations:
\begin{eqnarray}
\sum_{n,m=1}^\infty p_np_mn (1-s^n)(1-t^m) &\le& \sum_{n,m=1}^\infty p_np_m m(1-s^n)(1-t^m)\\
\sum_{n=1}^\infty \sum_{m=1}^\infty p_n p_mn (1-s^n)(1-t^m) &\le& \sum_{n=1}^\infty \sum_{m=1}^\infty p_n p_m m(1-s^n)(1-t^m)\\
\sum_{n=1}^\infty p_n n (1-s^n)\sum_{m=1}^\infty p_m (1-t^m) &\le& \sum_{n=1}^\infty p_n (1-s^n)\sum_{m=1}^\infty mp_m (1-t^m)\\
\frac{\sum_{n=1}^\infty p_n n (1-s^n)}{\sum_{n=1}^\infty p_n (1-s^n)} &\le& \frac{\sum_{m=1}^\infty mp_m (1-t^m)}{\sum_{m=1}^\infty p_m (1-t^m)}
\end{eqnarray}
which is what we wanted.

As an addendum, since it might not be obvious that $\frac{1-x^m}{1-x^n}$ is increasing. Note $$
\frac{d}{dx} \frac{1-x^m}{1-x^n} = \frac{mx^m(x^n-1) - n(x^m-1)x^n}{x(1-x^n)^2}
$$
which we can show is always non-negative for $x\in(0,1)$, $m \ge n$. The denominator is clearly positive for $x\in(0,1)$. Observe the numerator: $$
mx^m(x^n-1) - n(x^m-1)x^n = (m-n)x^{m+n} - m x^m + n x^n = x^n\left((m-n)x^m - m x^{m-n} + n\right)
$$
Obviously $x^n\ge 0$. The factor $(m-n)x^m - m x^{m-n} + n$ is decreasing as a function of $x$, so it achieves its minimum value at $x=1$, where it equals 0. Hence it is non-negative, and we conclude $\frac{d}{dx}\frac{1-x^m}{1-x^n}$ is always $\ge0$ for $x\in(0,1)$.
