What would be the nicest proof of the following theorem:

If $\lim\limits_{n \rightarrow \infty} a_{n} = \infty$, then $\lim\limits_{n \rightarrow \infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = e$.

If $\lim\limits_{n \rightarrow \infty}b_{n} = 0$, then $\lim\limits_{n \rightarrow \infty} \left(1 + b_{n} \right) ^ {\frac {1} {b_{n}} } = e$.

I somehow failed to find a proof here on the website and in the literature.

  • $\begingroup$ Use the fact $(1+1/n)^n$ is monotone. $\endgroup$ – Pedro Tamaroff Dec 6 '17 at 17:06
  • $\begingroup$ Are the $a_n$ and $b_n$ integers or real numbers? $\endgroup$ – Arthur Dec 6 '17 at 17:10
  • $\begingroup$ @Arthur If they were integers, the theorem would follow from the fact that the limit of any subsequence is in fact the limit of the whole sequence, I think. But they are not necessarily integers. $\endgroup$ – Theta Dec 6 '17 at 17:18
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    $\begingroup$ "What would be the nicest proof", asked 35 minutes ago ... answer already accepted. Hmmm ... I am sure there will be more people coming with various interesting proofs, but ... leave the question open for at least a week before choosing the "nicest" ;) $\endgroup$ – rtybase Dec 6 '17 at 17:38
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    $\begingroup$ @rtybase Haha, I should have added I meant nicest in terms of being elementary, then RRL's answer is hard to beat. :D $\endgroup$ – Theta Dec 6 '17 at 17:43


$$\left(1 + \frac{1}{\lfloor a_n \rfloor+1} \right)^{\lfloor a_n \rfloor} \leqslant \left(1 + \frac{1}{a_n} \right)^{a_n} \leqslant \left(1 + \frac{1}{\lfloor a_n \rfloor} \right)^{\lfloor a_n \rfloor+1}, $$


$$\left(1 + \frac{1}{n+1} \right)^n, \left( 1 + \frac{1}{n} \right)^{n+1} \to e$$

  • $\begingroup$ That's infact the same way you follow to demonstrate the basic limits, I don't really see any big insight in this. $\endgroup$ – user Dec 6 '17 at 17:32
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    $\begingroup$ Just a way to apply the squeeze theorem when $a_n$ is not an integer, given the limit result for $(1 + 1/n)^n$ where $n$ is an integer. $\endgroup$ – RRL Dec 6 '17 at 17:36
  • $\begingroup$ @RRL Thank you, a brilliant answer; somehow making $a_{n}$'s integers was what I was looking for (but didn't come up with a way to do this). $\endgroup$ – Theta Dec 6 '17 at 17:37
  • $\begingroup$ @Theta: You're welcome. $\endgroup$ – RRL Dec 6 '17 at 17:52

Given that $\lim_{n\to\infty }b_n = \lim_{n\to\infty }\frac{1}{a_{n}}=0$ we have,

$$\lim_{n \to\infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = \lim_{n \to\infty} \exp\left(\frac{\ln\left(1 + \frac{1}{a_{n}} \right)}{\frac1{a_{n}}} \right)= \lim_{h \to0} \exp\left(\frac{\ln\left(1 +h \right)}{h} \right)=e$$


$$\lim_{n \to\infty} \left(1 +b_n \right) ^ { \frac{1}{b_{n}}} = \lim_{n \to\infty} \exp\left(\frac{\ln\left(1 + b_n\right)}{b_n} \right)= \lim_{h \to0} \exp\left(\frac{\ln\left(1 +h \right)}{h} \right)=e$$


For $|t|<1$, note that $t-{1 \over 2} t^2 \le \log(1+t) \le t$ and so $|\log(1+t)-t| \le {1 \over 2} t^2$.

Hence for $|x|>1$ we have $|\log(1+{1 \over x})-{1 \over x}| \le {1 \over 2} ({1\over x})^2$ and so $|x\log(1+{1 \over x})-1| \le {1 \over 2} {1\over |x|}$.

Hence $\lim_{x \to \infty} x\log(1+{1 \over x}) =\lim_{x \to \infty} \log(1+{1 \over x})^x = 1$ from which it follows that $\lim_{x \to \infty} (1+{1 \over x})^x = e$.


One does L'Hospital to show that $\lim_{x\rightarrow\infty}\left(1+\dfrac{1}{x}\right)^{x}=e$, then one does sequential characterisation of limit. Similar reasoning applied to $b_{n}\rightarrow 0$ case.

  • $\begingroup$ As sequences are introduced earlier in a calculus course, supposse I don't have suffient knowledge on L'Hospital and functions in general, could you suggest a solution? $\endgroup$ – Theta Dec 6 '17 at 17:17
  • $\begingroup$ @RRL has provided a fantastic answer. $\endgroup$ – user284331 Dec 6 '17 at 17:23

For the function-sequence criteria, they are direct consequence of the two basic limits:

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

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

by a simple change of variables.


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

is obtained from the first by the substitution $$m=\frac1n$$


Proving the limit of a function of a sequence is equal to the function of the limit of that sequence

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    $\begingroup$ No, it's not a direct consequence by a simple change of variables. No one said anything about the $a_n$ being integers. $\endgroup$ – Arthur Dec 6 '17 at 17:10
  • $\begingroup$ @Arthur: If you read this answer carefully, he never required $n$ to be an integer. $\endgroup$ – Bumblebee Dec 6 '17 at 17:14
  • $\begingroup$ I'm referring to the function-sequence criteria. $\endgroup$ – user Dec 6 '17 at 17:15
  • $\begingroup$ When we read it carefully, the limits are not even right. $\endgroup$ – Raskolnikov Dec 6 '17 at 17:16
  • $\begingroup$ @Arthur You can also take a look here: math.stackexchange.com/questions/554910/… $\endgroup$ – user Dec 6 '17 at 17:18

Let's do the 1st one ...

One of the logarithmic inequalities says: $$\frac{x}{1+x} < \ln (1 + x) < x, \forall x > -1$$

Because $\lim\limits_{n\rightarrow \infty} a_n \rightarrow \infty$, then $a_n > 0$ and $\frac{1}{a_n} > 0$ from some $n$ onwards. So $$0<\frac{1}{a_n+1}=\frac{\frac{1}{a_n}}{1+\frac{1}{a_n}} < \ln\left(1 + \frac{1}{a_n}\right) < \frac{1}{a_n}$$ Given $e^x$ is ascending: $$1<e^{\frac{1}{a_n+1}} < 1 + \frac{1}{a_n} < e^{\frac{1}{a_n}}$$ and from some $n$ onwards $$e^{\frac{a_n}{a_n+1}} < \left(1 + \frac{1}{a_n}\right)^{a_n} < e$$ Squeeze theorem finishes the proof.


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