💾 Archived View for jsreed5.org › math › 20230123-the-limit-of-compound-interest › index.gmi captured on 2024-08-25 at 00:47:21. Gemini links have been rewritten to link to archived content
⬅️ Previous capture (2023-01-29)
-=-=-=-=-=-=-
Download a PostScript version of this document
(PS, 222 KB)
---
A common point of focus for analysis and finance students is the question of continual compound interest: how will a principal amount grow if an arbitrarily-small amount of compound interest in applied arbitrarily frequently? At the core of this question is the limit
lim((1+1/n)^n,n,infinity)=e
where e=2.71828... is Euler's number.
In many introductory courses, the above limit is given as a definition of the value of e. My own teachers never explained how to derive the identity. Several years later, a student of mine asked why the equation is true, and I set out to justify it for myself. The proof eventually came to me using calculus.
In this note, we assume we understand the properties of the logarithm function f(x)=log(x), where log(x) denotes the natural logarithm. We assume the value of e has been established via some other identity, such as e=∑(1/n!,n,0,infinity).
Suppose the value of the limit is an unknown constant a:
a=lim((1+1/n)^n,n,infinity)
We take the log of both sides:
log(a)=log(lim((1+1/n)^n,n,infinity))
The right-hand side is positive, and the logarithm function is continuous for all positive arguments, so it can be absorbed into the limit:
log(a)=lim(log((1+1/n)^n),n,infinity) =lim(n*log(1+1/n),n,infinity) =lim(log(1+1/n)/(1/n),n,infinity)
We now have a quotient of two continuous functions, both of which are greater than 0 for positive n: f(n)=log(1+1/n) and g(n)=1/n. Both of these functions approach 0 as n becomes arbitrarily large, giving us an indeterminate form. We thus need to use L'Hôpital's rule to evaluate the limit, which states (with the above conditions on f(n) and g(n)):
lim(f(n)/g(n),n,infinity)=lim((df/dn)/(dg/dn),n,infinity)
This gives us:
log(a)=lim(log(1+1/n)/(1/n),n,infinity)=lim((d/dn(log(1+1/n)))/(d/dn(1/n)),n,infinity) =lim(((-1/n^2)/(1+1/n))/(-1/n^2),n,infinity)
Simplifying:
log a=lim(((-1/n^2)/(1+1/n))(1/(-1/n^2)),n,infinity) =lim(1/(1+1/n),n,infinity) =1/(1+0) =1
Our final step is to clear the logarithm:
a=e^1=e
The fact that I was never exposed to a proof like this surprises me. In applied mathematics fields such as engineering and economics, many abstract statements are provided without proof. Applied fields do not need to know the underlying mechanics of the equations to trust in their correctness. Theoretical study, however, needs to ensure that no statement goes unjustified. Anything that can be proven must--the number of assumptions made should be reduced to a minimum.
Unfortunately, in practice, students are shown numerous statements that do not receive their own justification. Of course several curricula cover mathematical concepts that are necessary to know but not yet understand. But those concepts should be noted and expanded upon at the appropriate time. Otherwise students are inclined to accept statements based on the authority of the teacher or professor, which can lead to faulty leaps of logic when they need to write proofs on their own.
---
[Last updated: 2023-01-23]