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Consider the sequence $\{a_n\}$ defined as follows $$\{a_n\}_{n\ge{1}}=\left(1+\frac{1}{n}\right)^n\;,\; n\in\mathbb{N}$$ The question is to prove that $\{a_n\}$ has a limit as $n\to\infty$ and to find that limit.

This is a well known sequence and everbody in our class knows that the limit is Euler's number (i.e $e=2.718\ldots$). Indeed in high-school 'e' was defined as the limit of this particular sequence. However none of us could come up with a rigrous proof that this is indeed true.

Do note that since we have only touched upon limits so far in our calculus class, our teacher probably doesn't want a proof involving integrals.

Hosein Rahnama
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gandalfthegrey
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    "Indeed in high-school 'e' was defined as the limit of this particular sequence. However none of us could come up with a 'rigrous' proof that this is indeed true." If this is a definition, there is nothing to prove. – Taladris Aug 19 '14 at 10:15
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    One thing is to show that this series has a limit which should be no problem. The other is to find the limit. For this, the relevant question is if in your class you have already defined $\mathrm e$ and how you've defined it. If you did it using the series (as is usual), then one of the answers below will do. – Frunobulax Aug 19 '14 at 10:45

3 Answers3

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$$\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n=\lim_{n\to\infty }e^{n\ln\left(1+\frac{1}{n}\right)}=\lim_{n\to\infty }e^\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}=\lim_{x\to 0}e^{\frac{\ln(1+x)}{x}}.$$

Moreover $$\lim_{x\to 0}\frac{\ln(1+x)}{x}\underset{(1)}{=}\lim_{x\to 0}\frac{1}{1+x}=1$$

Then $$\lim_{x\to 0}e^{\frac{\ln(1+x)}{x}}\underset{(2)}{=}e^{\lim_{x\to 0}\frac{\ln(1+x)}{x}}=e^1=e$$

where $(1)$ holds by l'Hôpital's rule and $(2)$ by continuity of exponential function.

Hosein Rahnama
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idm
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    IF one has already defined a function $g$ that is inverse to the natural logarithm and one has already checked that one’s definition of the logarithm has $\ln(x+1)/x\to1$ and we accept $e:=g(1)$, then this is a good answer. However I recommend the Wikipedia article on equivalent characterisations of $e$ to any future reader wanting to explore definitions. – FShrike Dec 26 '22 at 15:22
  • @FShrike: Regarding your comment, I think that this post can be helpful for future readers. – Hosein Rahnama Dec 26 '22 at 15:27
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$a_n$ is increasing and bounded above by $3$, so converges. To show the limit equals to $e$, you can take $\ln a_n$,and show that $\ln a_n \to 1$. These are the steps you can take to prove it.

DeepSea
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You mean numerically? Basically you need to prove: $$ \lim_{n\to\infty}(1+\frac{1}{n})^n=\sum_{k=0}^\infty\frac{1}{k!} $$ And then the latter series gives you a way to compute (approximate value of) $e$, which is a symbol to denote the limit.

Troy Woo
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