Minimize series limit For a sequence $\{a_n\}$ of non-negative reals that are monotonically increasing, we want to minimize the following two expressions:
$$\sup_k\left(\frac{1}{a_{k-1}}\sum_{n=0}^ka_n\right)$$
$$\lim_{k\to\infty}\left(\frac{1}{a_{k-1}}\sum_{n=0}^ka_n\right)$$
Among sequences of the form $a_n=b^n$, we can explicitly compute the expression: $$\frac{1}{b^{k-1}}\sum_{n=0}^kb^n=\frac{1}{b^{k-1}}\frac{b^{k+1}-1}{b-1}\to\frac{b^2}{b-1}$$ which is minimized for $b=2$, so $a_n=2^n$ minimizes both the supremum and limit expressions.
I suspect $2^n$ achieves the minimum possible value for both expressions.  How could this be proven? Or how could I construct a sequence which achieves a lower value?
 A: This answer just clarifies my previous comment (since there is no need to consider $\beta$ as in my first comment): 
Consider a general sequence of positive nondecreasing numbers $\{a_n\}_{n=0}^{\infty}$.  Define 
$$\alpha = \limsup_{i\rightarrow\infty} a_{i}/a_{i-1}$$
So $1\leq \alpha \leq \infty$.  You can remove the case $\alpha = \infty$ since then $\sup_k \frac{1}{a_{k-1}}\sum_{n=0}^k a_n = \infty$.  Now for all suitably large $i$ we have that the ratio $a_i/a_{i-1}$ is "approximately" at most $\alpha$ (meaning the ratio is no more than $\alpha + \epsilon$ for any desired $\epsilon>0$).  So for suitably large indices $k$ for which $a_k/a_{k-1} \approx \alpha$ we get
$$ \frac{1}{a_{k-1}}[a_0+...+a_{k-2} + a_{k-1} + a_k] \approx  [stuff] + 1 + \alpha \geq ...$$
A: This is too long for a comment but the poster and I have "heuristically" shown that $a_k = 2^k$ must be optimal. I am not sure if the details are correct but hopefully someone can comment on it ? 
We can simplify our expression as 
$$ \frac{1}{a_{k-1}} \sum_{n=0}^{k} a_n = \frac{1}{a_{k-1}} \sum_{n=0}^{k-2} a_n + 1 + \frac{a_{k+1}}{a_k}.$$
If 
$$\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} \rightarrow L < \infty,$$
where $L > 1$ then we know that $a_k$ is $\sim CL^{k}$ for some constant $C$. The original poster has already posted an argument that if $a_k$ is geometric, then powers of $2$ are optimal.
Otherwise, if the ratio above converges, we must have 
$$\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} \rightarrow 1.$$
In this case, for $k$, we get that 
$$ \sum_{n=0}^k a_n = O(k) \cdot a_k \implies \frac{1}{a_k} \sum_{n=0}^k a_n = O(k).$$
Therefore, the original expression must diverge. Thus, powers of $2$ are optimal. However, as pointed out in the comments, the argument does not hold if the ratio does not converge. 
