A Generalization of Euler’s Limit

MSC:

• Primary 11Y60
• 40A05
• Secondary 11B83
• 11B05

Euler’s limit is defined as $\underset{n\to \infty }{\lim }{(\frac{n+1}{n})}^n=e$. We establish a generalization of this limit in the following proposition.

Proposition.

Let A_n be a strictly increasing sequence of positive numbers satisfying the asymptotic formula $A_{n+1}\sim A_n$, and let $d_n=A_{n+1}-A_n$. Then (1) $$\underset{n\to \infty }{\lim }{\left(\frac{A_{n+1}}{A_n}\right)}^{\frac{A_n}{d_n}}=e.$$(1)

Proof.

Let us consider the function $\text{ln}\hspace{0.17em}x$ on the interval $[A_n,A_{n+1}]$ for all $n\in \mathbb{N}$. By the mean value theorem, we have $\text{ln}\hspace{0.17em}{A}_{n+1}-\hspace{0.17em}\text{ln}\hspace{0.17em}{A}_n=\frac{1}{c}(A_{n+1}-A_n)$ for some c with $A_n<c<A_{n+1}$. Hence (since $\frac{1}{A_{n+1}}<\frac{1}{c}<\frac{1}{A_n}$)$$\frac{A_{n+1}-A_n}{A_{n+1}}<\hspace{0.17em}\text{ln}\hspace{0.17em}{A}_{n+1}-\hspace{0.17em}\text{ln}\hspace{0.17em}{A}_n<\frac{A_{n+1}-A_n}{A_n}.$$

Since $A_{n+1}\sim A_n$, we have$$1\leftarrow \frac{A_n}{A_{n+1}}<\frac{\hspace{0.17em}\text{ln}\hspace{0.17em}{A}_{n+1}-\hspace{0.17em}\text{ln}\hspace{0.17em}{A}_n}{\frac{A_{n+1}-A_n}{A_n}}<1;$$that is,$$\underset{n\to \infty }{\lim }\hspace{0.17em}\text{ln}\hspace{0.17em}{\left(\frac{A_{n+1}}{A_n}\right)}^{\frac{A_n}{A_{n+1}-A_n}}=1.$$

This completes the proof. ◼

It can be seen that generalization (1) gives Euler’s limit when A_n = n.