Bound exponent of a random variable Let $X$ be a random variable taking values in $[n] = \{1,\ldots,n\}$, such that $\mathbb{E}[X]= c$ for some $c$ (the exact distribution of $X$ is unknown).
Define $A = \sum_{i=1}^n \epsilon^i \Pr[X=i]$, for some $\epsilon<1$. I would like to bound $A$ somehow. A trivial bound is $ A \leq \sum_{i=1}^n \epsilon^i$, but I wonder if something better can be done. I noted that $A = \mathbb{E}[\epsilon^X]$, but not sure if it helps. 
 A: Define $A(x) = \sum_{k=1}^n x^k \Pr[X=k]$ (for $x\in[0,1]$).
By Jensen's inequality, you have, for $t \in\mathbb{R}$,
$$
\mathbb{E}[e^{t X}] \geq e^{t\mathbb{E}[X]}
$$
so you can already get (for $t=\ln\varepsilon$) a lower bound of 
$$\varepsilon^{\mathbb{E}[X]} \leq A(\varepsilon) \tag{1}.$$
Now, you want an upper bound. Unfortunately, without any knowledge of the distribution $(p_k)_{1\leq k\leq n}$ of $X$ (where $p_k := \Pr[X=k]$, you can't really hope for better than the $\approx \varepsilon $ you got (of course, you can get $\approx \varepsilon \lVert p\rVert_\infty$, but that's still linear in $\varepsilon$). 
To see why, assume with (some) loss of generality that $1<c\leq n/2$, and consider the distribution supported on two elements, 1 and $n$, with
$$
p_1  = \frac{n-c}{n-1}, \qquad p_n = 1- p_1 = \frac{c-1}{n-1}\,.
$$
Then $$\mathbb{E}[X] = 1\cdot p_1 + n\cdot p_n = n - (n-1)p_1 = c$$
but
$$
A(\varepsilon) = \varepsilon p_1+\varepsilon^n p_n  \geq \varepsilon p_1 \geq \frac{\varepsilon}{2} \tag{2}\,.
$$
So, at the very least, you can't really hope for a significantly better upper bound than the trivial one when $c \leq n/2$.
