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...
 A: Looking at this with no equations, a simple thought experiment. If, in fact, it grew without bounds, at some point, you have over 800kg. And with a 1 yr half-life, it will drop to 400kg a year later. Now, during that year, you've only added 365kg, and it's been there an average of 6 months so .707 remains (the square root of 'half'), say 30% gone or about 250 remaining. We now have just 650kg. 
The other answers clearly show precise numbers, but when I read your question, it seemed suited to a simple back of envelope approach to simply prove the negative. 
A: Suppose that at a certain stage we have an amount $M$ of the substance in the container. Then after $1$ day, the amount left is $M(1/2)^{1/365}$, so the amount that has died is $M(1-(1/2)^{1/365})$. If $M$ is large enough, the amount that has died in $1$ day is greater than $1$ kilogram. So the added $1$ kilogram cannot compensate for the amount that has died.
It follows that the amount of the substance in the container can never be greater than $M$. To be explicit, the amount cannot ever be greater than $\dfrac{1}{1-(1/2)^{1/365}}$.
A: Let's model this as a recursive sequence $X_1, X_2, \cdots, X_n, \cdots$, where $X_n$ is the number of kilograms of the element in the contains on the $n$th day.  Since the half life of the element is $1$ year, after a single day a proportion of $$\sqrt[365]{\frac{1}{2}}\approx 0.998$$
of the substance you began with will remain.  Since we also add $1$ kilogram after each day, the amount of the elment on the $n+1$st day will be
$$X_{n+1} = \sqrt[365]{\frac{1}{2}}X_n + 1$$
$$= \sqrt[365]{\frac{1}{2}}\left(\sqrt[365]{\frac{1}{2}}X_{n-1} + 1\right) + 1 = \frac{1}{2^{2/365}}X_{n-1} + \frac{1}{2^{1/365}} + 1$$
$$\vdots$$
$$= \frac{1}{2^{(n+1)/365}} + \frac{1}{2^{n/365}} + \cdots + \frac{1}{2^{1/365}} + 1$$
This is a geometric sum, which we can express as
$$X_{n+1} = \frac{1-2^{-(n+2)/365}}{1-2^{1/365}}.$$
So, substituting $n$ for $n+1$, 
$$X_n = \frac{1-2^{-(n+1)/365}}{1-2^{1/365}}.$$
Taking the limit as $n$ becomes very large, we find
$$\lim_{n\to \infty} X_n = \frac{1}{1-2^{1/365}} \approx 527,$$
so, after the process goes on for a very long time, the amount of radioactive material you have will be bounded, never growing to be much larger than $527$ (i.e., the exponential decay wins out over the arithmetic growth).
A: On the first day, you have 1 kilogram of decayium. On the second day, that 1 kilogram decays into $1/\sqrt[365]{2}$ kilograms of decayium, on the third day you have $1/(\sqrt[365]{2})^2$ kilograms from the original... And so on, until after a year has passed, you finally have $1/(\sqrt[365]{2})^{365} = 1/2$ kilograms form the original. Seems legit up until here.
But everyday, you're adding 1 kilogram to whatever already exists, so you have multiple copies of this sequence all running in parallel. So on the second day, there are $1 + 1/\sqrt[365]{2}$ kilograms, on the third day, $1 + 1/\sqrt[365]{2} + 1/(\sqrt[365]{2})^2$, ... and on the $n$th day, you will have
$$\sum_{i=0}^{n-1} \frac{1}{(\sqrt[365]{2})^i}$$
kilograms of decayium.
This is a geometric sequence, which we know has a sum of $\frac{1 - 1/(\sqrt[365]{2})^n}{1 - 1/\sqrt[365]{2}}$. As $n$ becomes larger and larger (tends to infinity), this actually converges: it approaches $\frac{1}{1 - 1/\sqrt[365]{2}} = 527$ (approximately). So it becomes a stable system of sorts, which means that you'll keep getting closer to 527 kilograms no matter how long you continue adding material.
A: 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.
