# Finding an Explicit Formula for a Geometric Series

CONTEXT: Uni question made up by lecturer

If you have to take a $$100$$mg drug every $$8$$ hours, and just before you take the drug, $$20$$% of it remains in your body, how would you write an explicit formula for this?

I've worked out that the recursive formula would be for integers $$n\ge0$$ where $$Q_0=100$$, I have $$Q_{n+1}=0.2Q_n+100$$ where $$Q_n$$ is the quantity of the drug in the body just after the $$n$$th dose is taken.

My lecturer gave us a hint saying the solution involves geometric series, but I'm not sure how this can be the case since the recursive definition I've developed involves an added constant of $$100$$.

The explicit definition is then to be used to find the quantity of the drug remaining in the body in the long run (which would just be what the series converges to).

Imagine building it up iteratively, generally factoring out $$100$$ whenever possible: you should get,

• $$Q_1 = 100$$
• $$Q_2 = 0.2 Q_1+ 100 = 0.2 \cdot 100 + 100 = 100(1+0.2)$$
• $$Q_3 = 0.2Q_2 + 100 = 0.2(100(1+0.2)) + 100 = 100(1+0.2+0.2^2)$$
• $$Q_4 = 0.2Q_3 + 100 = 100(1+ 0.2 + 0.2^2 + 0.2^3)$$

Then visibly, in general,

$$Q_n = 100 \sum_{k=0}^{n-1} 0.2^k$$

This can be simplified with the formula for a finite geometric series.

• How come $Q_3 = 0.2(100(1+0.2)+100)+100$? Shouldn't it be $Q_3 = 0.2(100(1+0.2))+100$? Commented May 4, 2019 at 7:48
• Yeah, it should. I went back and forth between a few ways of communicating the method and I guess some typos resulted. My bad. Commented May 4, 2019 at 7:49

Good work so far: the recursion you've written is exactly correct. You can actually find the quantity of drug remaining in the body long-term using this and nothing more. To wit, notice that $$\displaystyle \lim_{n \to \infty} Q_{n+1} = \displaystyle \lim_{n \to \infty} Q_n = L$$ for some $$L \in \mathbb{R}$$, provided this limit actually exists$$^\dagger$$. Therefore, taking a limit on both sides of your equation gives:

$$L = 0.2L + 100$$

Solving for $$L$$, we find $$L = 125$$, which agrees with experimental results and Eevee's approach using geometric series.

$$^\dagger$$You can justify that the limit exists by applying the monotone convergence theorem after showing that the sequence is both increasing and bounded above.

• I probably would've just solved the recurrence directly, if the hint about geometric series weren't there, but that might be because I'm fresh out of a combinatorics class that featured them a lot. This seems a hell of a lot simpler than both methods though. xD Commented May 4, 2019 at 7:42

We can use Generating functions to solve the recurrence relation. Indeed put $$Q(z)=\sum_{n=0}^\infty Q_{n}z^n\tag{0}$$ (it converges in a small neighbourhood of the origin by the recurrence relation) so that the recurrence $$Q_{n+1}=0.2Q_n+100;\quad Q_0=100$$ implies that $$\frac{Q(z)-100}{z}=0.2Q(z)+\frac{100}{1-z}.$$ Solve for $$Q(z)$$ to get that $$Q(z)=\frac{100}{(1-z)(1-0.2z)}=100\times\frac{1}{1-z}\times \frac{1}{1-0.2z}\tag{1}.$$ At this point we can proceed to use partial fraction expansion or we can use convolution. In general if $$A(z)=\sum_{n=0}^\infty a_n z^n;\quad B(z)=\sum_{n=0}^\infty b_nz^n$$ then $$A(z)B(z)=\sum_{n=0}^\infty\left(\sum_{k=0}^na_kb_{n-k}\right)z^n \tag{2}.$$ From $$(1)$$ put $$(1-z)^{-1}=\sum_{n=0}^\infty z^n=B(z)$$ and $$(1-0.2z)^{-1}=\sum_{n=0}^\infty 0.2^n z^n=A(z)$$ into $$(2)$$ so that $$\frac{1}{1-z}\times \frac{1}{1-0.2z}=\sum_{n=0}^\infty\left( \sum_{k=0}^n (0.2)^k \right)z^n$$ whence $$Q(z)=\sum_{n=0}^\infty Q_{n}z^n=100\times\frac{1}{1-z}\times \frac{1}{1-0.2z}=\sum_{n=0}^\infty\left( 100\sum_{k=0}^n (0.2)^k\right)z^n.$$ So $$Q_n=100\sum_{k=0}^n (0.2)^k\quad (n\geq 0).$$