3

Use mathematical induction to prove that for any $k \in \mathbb N, \lim (1+k/n)^n = e^k$.

I already used monotone Convergence Theorem to prove $k=1$ case. Do I just need to go through the same process to show $k$? If not, could you please help?

Thanks

afsdf dfsaf
  • 1,737

3 Answers3

5

First a useful bound: Suppose $\theta_n \ge 0$ is such that $\lim_n n \theta_n = 0$, then $\lim_n (1+ \theta_n)^n = 1$. To see this, note $(1+\theta_n)^n = \sum_{k=0}^n \binom{n}{k} \theta^k \le \sum_{k=0}^n n^k \theta^k \le \frac{1}{1-n \theta_n}$.

Suppose the result is true for $k$ (I am taking $k=1$ as already proved). Then note that $(1+ \frac{k}{n}) (1+ \frac{1}{n}) = (1+ \frac{k+1}{n}+ \frac{k}{n^2}) = (1+ \frac{k+1}{n})(1+ \frac{k}{n^2 (1+ \frac{k+1}{n})}) $, and let $\theta_n = \frac{k}{n^2 (1+ \frac{k+1}{n})}$. This gives $ (1+ \frac{k+1}{n})^n = \frac{1}{(1+\theta_n)^n} (1+ \frac{k}{n})^n (1+ \frac{1}{n})^n$. Taking limits gives $ \lim_n (1+ \frac{k+1}{n})^n = \frac{1}{1}e^k e^1 = e^{k+1}$, hence the result is true for $k+1$.

This was my original answer which is not correct for the reason Robjohn points out below:

Assuming that you have proved it for $k=1$, note that the function $x \mapsto x^k$ is continuous, and $(1+ \frac{k}{n})^n = (1+ \frac{k}{n})^{\frac{n}{k}k} = \left ( (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k$.

Hence $\lim_n (1+ \frac{k}{n})^n = \lim_n \left ( (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k = \left( \lim_n (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k = \left( \lim_n (1+ \frac{1}{\frac{n}{k}} )^{\frac{n}{k}} \right)^k = e^k $.

copper.hat
  • 178,207
  • Note that $\left(1+\frac1n\right)^n$ is a subsequence of $\left(1+\frac kn\right)^{n/k}$. Although the convergence of the sequence implies the convergence, to the same limit, of any subsequence, a bit more is needed to show the convergence of the sequence from the convergence of a subsequence. However, I like your alternative approach (as long as $\theta_n\ge0$) (+1). – robjohn Oct 23 '13 at 10:46
  • @robjohn: Good point, it didn't even occur to me. I will make the alternative my answer instead. – copper.hat Oct 23 '13 at 15:05
  • Another proof, that works for negative $\theta_n$ as well, uses Bernoulli's inequality: $$ 1+n\theta_n\le(1+\theta_n)^n\le\frac1{(1-\theta_n)^n}\le\frac1{1-n\theta_n} $$ Then use the squeeze theorem. – robjohn Oct 23 '13 at 15:23
  • @robjohn: My original alternative was an afterthought since it occurred to me that I hadn't actually answered the actual question ('use induction'). A nicer bound would have been $|(1+\theta_n)^n-1| \le \frac{|n \theta_n|}{1-|n \theta_n|}$. (I thought I had found a way around showing the sequence was monotonic with the first approach.) – copper.hat Oct 23 '13 at 15:31
4

Suppose $$ \lim_{n\to\infty}\left(1+\frac1n\right)^n=e $$ Bernoulli's Inequality (which is proven by induction at the end of this answer) yields $$ \begin{align} \frac{\left(\color{#C00000}{1+\frac k{n+1}}\right)^{n+1}}{\left(\color{#00A000}{1+\frac kn}\right)^n} &=\color{#00A000}{\frac{n+k}{n}}\left(\color{#C00000}{\frac{n+k+1}{n+1}}\color{#00A000}{\frac{n}{n+k}}\right)^{n+1}\\ &=\frac{n+k}{n}\left(1-\frac{k}{(n+1)(n+k)}\right)^{n+1}\\ &\ge\frac{n}{n+k}\left(1-\frac{k}{n+k}\right)\\ &=1 \end{align} $$ That is, $\left(1+\frac kn\right)^{n/k}$ is increasing in $n$.

Since $\left(1+\frac 1n\right)^{n}$ is a subsequence of the increasing sequence $\left(1+\frac kn\right)^{n/k}$, they both tend to the same limit. That is, $$ \lim_{n\to\infty}\left(1+\frac kn\right)^{n/k}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e $$ which implies $$ \lim_{n\to\infty}\left(1+\frac kn\right)^n=e^k $$

Community
  • 1
robjohn
  • 353,833
  • Since I only show (1+1/n)^n converges to e and (1+1/n)^n is the subsequence of sequence (1+k/n)^(n/k). Based on Thm, if (1+k/n)^(n/k) converges t e, then its subsequence also converge to e. However, not the converse. – afsdf dfsaf Oct 24 '13 at 14:22
  • @afsdfdfsaf: Yes, but if a sequence is monotone, and a subsequence converges, then the sequence converges to the same limit. – robjohn Oct 24 '13 at 15:02
0

A similar yet different approach: Assume that $\lim_{n\to\infty}\left(1+\frac kn\right)^n=e^k$ for some $k$ and prove it for $k+1$. Notice that $$\begin{align} 1+\frac{k+1}{n} &= \frac{n+k+1}{n} \\ &=\frac{n+k+1}{n+k}\cdot \frac{n+k}{n} \\ &=\left(1+\frac{1}{n+k}\right)\cdot\left(1+ \frac{k}{n}\right) \end{align}$$ Now, $$\begin{align} \left(1+\frac{1}{n+k}\right)^n &=\frac{\left(1+\frac{1}{n+k}\right)^{n+k}}{\left( 1+\frac{1}{n+k}\right)^k} \end{align}$$
The numerator is a subsequence of $(1+\frac{1}{n})^n$ and therefore converges to $e$, and the denominator converges to $1$. Then the fraction tends to $e$ when $n\to\infty$. From limit arithmetic, we have $$\begin{align} \lim_{n\to\infty}\left(1+\frac{k+1}{n}\right)^n &=\lim_{n\to\infty}\left(1+\frac{1}{n+k}\right)^n\cdot\left(1+ \frac{k}{n}\right)^n \\ &=e\cdot e^{k} \\ &=e^{k+1}. \end{align}$$

Roy Sht
  • 1,429