Limit of a series with upper bound in the summand? I have constructed a model of drug dosing, and to find the maximum quantity of drug in the body after an infinite number of doses, I believe I must compute this limit:
$$\lim_{n \to \infty} D \displaystyle\sum_{i=1}^{n} e^{-(n-i)k\Delta t}$$
where $D, k, \text{and } \Delta t$ are constants. (The fixed dose amount, rate constant of idealized metabolism, and fixed interval between doses in my model.) I'm aware it can be reworked into the classic indeterminate forms $\frac{\infty} {\infty}$ and $0\times\infty$. I've played around with it a lot, and at one point concluded it equals $D/{(1-e^{-k\Delta t})}$, but I think there was a flaw in my derivation.
In general, how does one evaluate the following expression? 
$$\lim_{n \to \infty} \displaystyle\sum_{i=a}^{n} f(n-i)$$
(Assuming the summand function is of a form that will allow convergence, such as my $e^{-c(n-i)}$.)
Thank you for your wisdom!

Your sum is equivalent to $$\lim_{n\to \infty} De^{-nk\Delta t}\sum_{i=1}^{n} (e^{k\Delta t})^i$$
Which is a geometric series.

$$\lim_{n\to \infty} De^{-nk\Delta t}\sum_{i=1}^{n} (e^{k\Delta t})^i=\lim_{n\to \infty} De^{-nk\Delta t} \frac{e^{k\Delta t}(e^{k\Delta t n}-1)}{e^{k\Delta t}-1}$$
Carry out the multiplication to get
$$\lim_{n\to \infty} D\frac{e^{k\Delta t}(1-e^{-nk\Delta t})}{e^{k\Delta t}-1}$$
Since $$\lim_{x\to \infty}e^{-x}=0$$
You get
 $$D\frac{e^{k\Delta t}}{e^{k\Delta t}-1}$$
