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Since $$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n={e}$$

My strong hunch is that the following statement must also be true $$\lim_{n\rightarrow \infty}\left(1+\frac{r}{n}\right)^n = {e^{r}}$$ for all $r>0$.

But I can neither prove or disprove it, any idea on how to prove it? Or if the statement is not true, how it should be modified so that it is true?

Derek Allums
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Graviton
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2 Answers2

39

Your hunch is correct. Letting $u = \frac{n}{r}$, we have: $$\begin{align*} \lim_{n\to\infty}\left(1 + \frac{r}{n}\right)^n &= \lim_{n\to\infty}\left(\left(1+\frac{r}{n}\right)^{n/r}\right)^r\\ &= \lim_{u\to\infty}\left(\left(1 + \frac{1}{u}\right)^u\right)^r\\ &= \left(\lim_{u\to\infty}\left(1 + \frac{1}{u}\right)^u\right)^r\\ &= e^r. \end{align*}$$

Arturo Magidin
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  • I will add this proof but its a bit late. +1. – GarouDan Mar 06 '12 at 20:06
  • this prove requires the fact that $\lim_{u \to \infty, u \in \mathbb{R}} (1+1/u)^u = e$ to make the substitution since $n/r$ is not always an integer. The definition of $e$ only shows that $e$ is $\lim_{n \to \infty, n \in \mathbb{N}} (1+1/n)^n = e$. – juekai Jun 14 '24 at 15:58
  • @juekai "prove" is the verb, "proof" is the noun. As to the limits, they are equal, as the real limit function interpolates monotonically between the sequential values. Either limit establishes the other limit. – Arturo Magidin Jun 14 '24 at 16:33
16

Another way to see this. Suppose $$\lim_{n\to \infty} \left(1+\frac{r}{n}\right)^n = L.$$ Let us calculate $\ln(L)$:

$$\begin{align*} \ln(L) &= \ln\left(\lim_{n\to \infty} \left(1+\frac{r}{n}\right)^n \right)\\ &=\lim_{n\to \infty} \ln\left(\left(1+\frac{r}{n}\right)^n\right)\\ &=\lim_{n\to \infty} n\ln\left(1+\frac{r}{n} \right)\\ &=\lim_{n\to \infty} \frac{\ln\left(1+\frac{r}{n} \right)}{\frac{1}{n}}\\ &=\lim_{n\to\infty} \frac{\frac{1}{1+\frac{r}{n}}\cdot\frac{-r}{n^2}}{-\frac{1}{n^2}}\\ &=\lim_{n\to\infty} \frac{r}{1+\frac{r}{n}}\\ &=r, \end{align*}$$ where we have used the fact that $\ln(x)$ is continuous in $(0,\infty)$, and l'Hôpital's rule. Thus, $\ln(L)=r$, or equivalently, $L=e^r$.

Álvaro Lozano-Robledo
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