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!


1 Answer 1


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 $$D\frac{e^{k\Delta t}}{e^{k\Delta t}-1}$$

  • $\begingroup$ 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. $\endgroup$ Dec 14, 2016 at 2:14
  • $\begingroup$ @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 $\endgroup$
    – GuPe
    Dec 14, 2016 at 2:42
  • $\begingroup$ 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. $\endgroup$ Dec 14, 2016 at 4:53
  • $\begingroup$ @electronpusher because while the sum grows, you are multiplying it by a term that is going to zero, so they balance out and converge. $\endgroup$
    – GuPe
    Dec 14, 2016 at 13:17
  • $\begingroup$ I agree intuitively, but neither us have proven that. Not every product of something growing to infinity times something going to zero results in a finite number. The product could also result in infinity or zero. This is the nature of the indeterminate limit form 0 times infinity, that I mentioned in the second paragraph of my OP. It doesn't matter anyway because you used the sum formula for a geometric series which, in the form you wrote it, does not converge. It is invalid to use the sum formula when the common ratio (under the index exponent) is greater than one. Please review. $\endgroup$ Dec 15, 2016 at 4:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .