How fast does a sequence with finite sum go to zero? Suppose $z_n$ is a nonnegative sequence, monotonically decreasing to zero, and $$\sum_{i=1}^{\infty} z_i < 1.$$ Is it possible to translate this into a bound on the how small $z_i$ is? For example, it seems natural to guess that $z_i < z_0 \alpha^i$ for some $\alpha$ that depends on $z_0$. Is this true? If not, can anything about the decay of the sequence be inferred from the sum bound?
 A: Consider the closely related problem where $\int_0^\infty f(x)\,dx<1$, with $f$ decreasing and positive. I find it easier to explain why you cannot accomplish this here than with series. But the idea can be adopted.
You could always substitute $x=\log(u+1)$ (preserving area) to get $\int_0^\infty \frac{f(\log(u+1))}{1+u}du$, and now you have a function that decreases substantially more slowly than $f$ does. (The substitution is pushing the bulk of the area under the curve far to the right.) If you had a bound, something like this will break it. And if this isn't enough to break it, you can replace $\log(u+1)$ with slower and slower concave down increasing functions, like $\log(\log(\log(u+1)+1)+1)$.
A: At the very least, such an $\alpha$ might not exist.  As an example, consider $z_i = 1/i^{1+\epsilon}$ for any (small) $\epsilon > 0$.
Such an $\alpha<1$ exists exactly when $\sum z_i$ can be shown to converge by the ratio-test.
A: Well, since it's monotonically decreasing, $z_n\leq z_{n-1}$. This means that 
$nz_n \leq \sum_{i=1}^n z_i \leq \sum_{i=1}^{\infty} z_i < 1$
so $z_n < {1 \over n}$.
