Derive e

Prove that \( \left( 1 + \frac{1}{n} \right)^n \) converges.

Solution

Let \( G(n) =  \left( 1 + \frac{1}{n} \right)^n \).

Binomial expansion, \( G(n) = \sum_{k = 0}^{n} \frac{ \binom{n}{k} }{n^k} \).

By simple inequality argument, \( \frac{ \binom{n}{k} }{n^k} < \frac{1}{k!} \).

So if we let \( F(n) = \sum_{k = 0}^{n} \frac{1}{k!} \), we have \( G(n) < F(n) \).

By simple inequality argument, \( F(n) < 1 + \sum_{k = 1}^{n} \frac{1}{2^{k-1}} < 3 \).

So G and F are bounded above.

By AM-GM, \( \frac{G(n-1)}{G(n)} = \left( \frac{ n^2 }{ n^2 - 1} \right)^{(n-1)} \frac{ n}{n+1} < \left(  \frac{ \frac{ n^2 }{ n^2 - 1} (n-1) + \frac{ n}{n+1} }{n} \right)^n = 1 \). So \( G(n-1) < G(n) \).

Since G is monotonic increasing and bounded above, G converges.

Similarly F converges.

Let's also prove that G and F converges to the same limit.

By studying individual terms in \( F(n) - G(n) \) and making simple inequality arguments, we can show that for each \( \epsilon > 0 \) we can pick N s.t. \( F(n) - G(n) < \epsilon \) for all n > N. This can be further used to make the formal argument that the limit of F is the same as the limit of G.

So if we call the limit e, then \( \lim_{n \to \infty } \left( 1 + \frac{1}{n} \right) ^n  = 1 + \frac{1}{1!} + \frac{1}{2!} +  \frac{1}{3!} + \cdots = e \).