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I am in an intro to Analysis class, and I want to show that

$$\lim_{n\to \infty} \left( 1+\frac{1}{2n} \right)^n = \sqrt{e}$$

I already have a result that

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

Which was not proven rigorously but was given to us. We basically just proved that the limit was bounded between 2 and 3, and then they gave us the real limit $e$. I am not sure if I need to use that result or not.

I have also found in the text a property that says a suite of real numbers converges to $\sqrt{a}$ if $x_1>0$ and $x_n=\frac{1}{2}\left(x_{n-1}+\frac{a}{x_{n-1}}\right),n\ge2$

I am having trouble representing the first expression in this way. Any direction would be greatly appreciated.

jeffery_the_wind
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4 Answers4

12

We have that

$$\left( 1+\frac{1}{2n}\right)^n=\left[\left( 1+\frac{1}{2n}\right)^{2n}\right]^\frac12$$

then let $k=2n \to \infty$.

user
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  • This is very easy :-). +1 for all the users. – Sebastiano Oct 07 '19 at 16:16
  • Yes thanks for the quick explanation, but I am in an intro class and we are expected to show every part of the logic. I see what you did here but I am missing that part that we are allowed to assume that the limit of a product of a suite approaches the product of the limits of the component suites. – jeffery_the_wind Oct 07 '19 at 16:27
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    OK I guess I found it here: if $\lim_{n\to \infty} x_n = x$ and $\lim_{n\to \infty} y_n = y$ then $\lim_{n\to \infty} x_ny_n = xy$ – jeffery_the_wind Oct 07 '19 at 16:30
  • @jeffery_the_wind Yes of course! Refer to these notes. – user Oct 07 '19 at 16:43
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Hint:$$\left(1+\frac1{2n}\right)^n=\sqrt{\left(1+\frac1{2n}\right)^{2n}}.$$

José Carlos Santos
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5

$$ \left(1+\frac 1{2n}\right)^{2n}\longrightarrow e. $$ Now, by the continuity of the square root, $$ \left(1+\frac 1{2n}\right)^{n} = \left[\left(1+\frac 1{2n}\right)^{2n}\right]^{1/2}\longrightarrow \sqrt e. $$

amsmath
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0

There is a striking similarity between

$$\left(1+\frac1{n}\right)^n$$ and $$\left(1+\frac1{2n}\right)^n,$$

which you should try to exploit.

With a little bit of thinking and $m:=2n$, $$\left(1+\frac1{2n}\right)^n=\left(1+\frac1{2n}\right)^{2n/2}=\sqrt{\left(1+\frac1{m}\right)^m}.$$

And the limit works because the square root is a continuous function.