Proof that $e$ is finite

Question

Define $e-1$ as the yearly compound interest obtained from a one dollar investment with 100% gross annual interest:

$$ e := \lim_{n\to\infty} (1+1/n)^n =: \lim_{n\to\infty} a_n $$

Nine year olds can understand this, if it is explained using words instead of the limit symbol: "you can get 100% at the end of the year or you can get 50% after half a year and 50% of your updated account after another half year. Clearly the latter is better for you, so let's continue ..."

In fact, a nine year old I tested this with correctly concluded that $$ e>2 $$ since he intuited that $a_n$ is monotonic, which is obvious in the $1\to 2$ case and which is very easy to argue in the $2^n \to 2^{n+1}$ by repeating the argument on each subperiod.

However, I wasn't able to provide any argument that $e$ is even finite that doesn't require arithmetic manipulations on a sheet of paper. Two approaches I have seen are (1) (a) expand the definition of $a_n$ using the binomial theorem (b) bound the terms by 1 plus a geometric sequence or (2) prove that $b_n:=(1+1/n)^{n+1}$ is decreasing.

I feel like the latter might be just what I am looking for, if there is an intuitive argument that $b_n$ is decreasing.

Does anyone here know of a good intuitive argument or visual proof that the compound interest is finite?

Answer 1

I would base the 9-year old accessible argument on the following idea:

Let's look at how many multiplications by $1+\frac 1n$ are needed to fo from $1$ to $2$. Note that multiplication by $1+\frac 1n$ is the same as adding $1/n$-th of the quantity to the quantity. Suppose we need $m$ steps to reach or exceed $2$ for the first time. Then at each step we started with a quantity below $2$, so we added at most $2/n$. Hence $m\frac 2n\ge 2-1=1$, so $m\ge\frac n2$. Thus $(1+\frac 1n)^{\frac n2-1}\le 2$ and $(1+\frac 1n)^n\le 4(1+\frac 1n)^2\le 16$, say.

Another possibility based on what you talked about already is to look at what happens if we double $n$ in $(1-\frac 1n)^n$. We have $$ \left(1-\frac 1{2n}\right)^{2n}=\left[\left(1-\frac 1{2n}\right)^2\right]^n=\left(1-\frac 1n+\frac1{4n^2}\right)^n\ge \left(1-\frac 1{n}\right)^{n}\,, $$ so if you restrict yourself to the subsequence $n=2^k$, you'll get $$ \left(1-\frac 1n\right)^n\ge \frac 14 $$ for $n\ge 2$.

But $(1+\frac 1n)^n(1-\frac 1n)^n=(1-\frac 1{n^2})^n\le 1$, so $e\le 4$.

Answer 2

Intuitively and visually, comparing the graphs of the functions

$$ y = (1+1/x)^x \;\;\mathrm{and}\;\;y = 2.72 $$

seems pretty convincing to me, and I expect would also be for a young math student. I would invite them to graph both functions using a tool like Desmos, then draw their own conclusion.

Answer 3

Cool question. I think maybe you're making the problem harder by talking about $e$ rather than by talking about the exponential function. If I define $\exp(x)=\lim (1+x/n)^n$, then the question is whether this function blows up to infinity before you get to $x=1$. Talk to the kid about yeast growing in a pot or whatever, and I think it's pretty easy to get them to the point where they understand that this is a description of a function that grows at a rate that's proportional to its current value. But for a function whose rate of growth is proportional to its current value, you're going to have a scaling property where every piece of the graph looks like every other piece of the graph, but just vertically scaled by some factor. That means that function can't blow up at $x=1$ unless it also blows up for positive values of $x$ that are as small as you like. But I think it's quite easy to convince the kid (at age-9 levels of rigor) that $\lim (1+0.01/n)^n$ is not infinite and is in fact very close to 1.01.

Answer 4

We interpret $(1+1/n)^n$ as the future value of one dollar invested at 100% for 1 year, compounded $n$ times. The OP is asking for an intuitive argument that this future value is not infinite if we compound continuously, i.e., if we let the number of compoundings $n$ per year go to infinity.

For intuition for why this future value is finite, let's tinker with the 1 year part of the model. Focus on the correct 1 in the formula: $(1+{{\bf 1}/n})^n$.

Maybe the year is a year on Mercury, which is roughly a quarter of an Earth year. If we do our calculations on Mercury, we would not need to wait an entire Earth year for our infinite payout. We would get this payout after just 3 months, i.e., one year on Mercury. Very odd.

Using imaginary worlds with years that are vanishingly short relative to an Earth year, we are led to live with a reality where $(1+{x}/n)^{n}$ tends to infinity for any $x>0$, where $x$ is the length of the year on the imaginary planet relative to our Earth year.

This would be a catastrophe for mathematics, because $$\exp(x)=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n$$ would be infinite for $x>0$.