# 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 $$D\frac{e^{k\Delta t}}{e^{k\Delta t}-1}$$
• Thank you, I am actually aware of that form. The parameters $k$ and $/Delta t$ are positive so the series written that way will not converge; we would be left with the $0 /times /infinty$ indeterminate form I alluded to. I am confident based on my model that this limit does indeed exist, but I have yet to obtain it by a rigorous method. Dec 14, 2016 at 2:14
• @electronpusher I added some details, in fact, because $k$ and $\Delta t$ are positive, we find that it does converge. Actually, as long as they have the same sign, the sum will converge
• Thank you for the clarification. Our answers agree. Can you please explain why you say this geometric series converges? I was under the impression the magnitude of the ratio, $e^{k /Delta t}$, must be less that unity otherwise the series will diverge. Dec 14, 2016 at 4:53