I'm trying to prove that if $x_n$ is a sequence such that $x_n \rightarrow +\infty$ then $(1+\frac{1}{x_n})^{x_n}\rightarrow e$. Here is what I've done so far:

We know that $\{(1+\frac{1}{n})^n\}_{n \geq 1} \rightarrow e$ so it's true that $\forall_{\epsilon >0}\exists_{n\geq 1}\forall_{n'\geq n}|(1+\frac{1}{n})^n-e|<\epsilon$. We will use that to pick $n'$ for a given $\epsilon$ later.

Now we know that $x_n \rightarrow +\infty$, so it is true that $\forall_{\epsilon > 0}\exists_{n' \geq n_0}\forall_{n\geq n'} x_n \geq \epsilon$.

Let's analyze $(1+\frac{1}{x_n})^{x_n}$ now. We want to show that it tends to $e$ so we're trying to show that

$$\forall_{\epsilon >0}\exists_{n'\geq n_0}\forall_{n\geq n'}|(1+\frac{1}{x_n})^{x_n}-e|<\epsilon$$

So we pick any $\epsilon$. And now we know that for sufficiently large $n$ (let's say $n\geq n'$) $|(1+\frac{1}{n})^n-e| < \epsilon$. And also for any given number we can make $x_n$ bigger than it so let's pick $n''$ such that $\forall_{n\geq n''} x_n > n'$. And that's it - for a given $\epsilon$ we've found $n''$ such that $\forall_{n\geq n''}|(1+\frac{1}{x_n})^{x_n}-e|<\epsilon$, which proves that $(1+\frac{1}{x_n})^{x_n}$ tends to $e$.

So is my solution correct? Also - in the workbook there was a hint that we should use squeeze theorem and the property that if a sequence has a limit then for all its subsequences their limit is the same but I don't see that. Does that lead to a better solution?

  • $\begingroup$ Your solution is not correct. Maybe take a look at how you prove that $(1+1/n)^n\to e$ and try from there. $\endgroup$ Jan 12, 2015 at 22:22
  • $\begingroup$ What is wrong with it? $\endgroup$
    – qiubit
    Jan 13, 2015 at 0:44
  • $\begingroup$ Is $x_n$ an integer? $\endgroup$ Jan 13, 2015 at 2:24
  • $\begingroup$ Not necessarily $\endgroup$
    – qiubit
    Jan 13, 2015 at 12:10

1 Answer 1


Your proof is much complicated. A simple and short argument is desirable.

If all of $x_n $ are integers, there is nothing to prove. We assume that $(x_n)$ is a sequence of real numbers. Take a sequence $(m_n)$ of integers with $m_n\le x_n< m_n+1$ for $n=1, 2,...$ . Since $$1+\frac{1}{m_n+1}<1+\frac{1}{x_n}\le 1+\frac{1}{m_n},$$ we have $$\left(1+\frac{1}{m_n+1}\right)^{m_n}<\left(1+\frac{1}{x_n}\right)^{x_n}\le \left(1+\frac{1}{m_n}\right)^{m_n+1}.$$ Here $(m_n)$ is a subsequence of $(n)_{n\ge 1}$, and hence we have $\lim_{ n\to \infty} \left(1+\frac{1}{m_n}\right)^{m_n}=e$.

Thus we have $$\lim_{ n\to \infty} \left(1+\frac{1}{m_n+1}\right)^{m_n}=\lim_{ n\to \infty} \frac{\left(1+\frac{1}{m_n+1}\right)^{m_n+1}}{\left(1+\frac{1}{m_n+1}\right)}=e$$ and $$\lim_{ n\to \infty} \left(1+\frac{1}{m_n}\right)^{m_n+1}=\lim_{ n\to \infty} \left(1+\frac{1}{m_n}\right)^{m_n}\left(1+\frac{1}{m_n}\right)=e.$$ Using the squeeze theorem, we have $ \lim_{ n\to \infty} \left(1+\frac{1}{x_n}\right)^{x_n}=e$.


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