Arithmetic growth versus exponential decay I have a kilogram of an element that has a long half-life - say, 1 year - and I put it in a container. Now every day after that I add another kilogram of the element to the container.
Does the exponential decay eventually "dominate" or does the amount of the substance in the container increase without bound?
I know this should be a simple answer but it's been too long since college...

Instead of looking at the whole sequence, you can look at a difference equation. Suppose $m(t)$ is the mass you after $t$ days. Put $\Delta(m)(t)=m(t+1)-m(t)$. What you are asking is, essentially, what is the (approximate) limit for a solution to the initial value problem
$$
\begin{cases}
 \Delta(m)=-(1-(\frac{1}{2})^{1/365})m+1\\
 m(0)=1
\end{cases}
$$
Consider the difference equation $\Delta(m)=-(1-(\frac{1}{2})^{1/365})m+1$. This is equal to zero precisely when $m$ is equal to the equilibrium mass $m_e=1/(1-(\frac{1}{2})^{1/365})\approx 527.083$, and moreover, whenever $m<m_e$ we have $\Delta(m)>0$, while whenever $m>m_e$ we have $\Delta(m)<0$. Moreover, if $m$ is very close to $m_e$, the absolute value $\lvert \Delta(m)\rvert$ is very small (so once we are close to $m_e$ we cant jump too far away), while it is relatively large while we are far away (so we can't suddenly slow down far away from $m_e$, so we will eventually get close).
This tells us that regardless of the initial value $m(0)$, the masses in consecutive days will either a) get closer and closer to $m_e$ indefinitely or b) get closer and closer and eventually it will go past $m_e$, and immediately start going back, perhaps overshooting $m_e$ again, possibly repeating the pattern indefinitely, and possibly only a few times.
If you were to add the new mass continually as opposed to every day, you would always end up in the case a) instead, and possibly you can rule out b) by analysing the numbers more carefully. Either way, no matter where you start (i.e. even if you start with a huge amount of the stuff, or somehow have it in the negatives to begin with), you will eventually get close to $m_e$ and stay close.
