I have to prove that the following limit exists: $$\lim_{n\to \infty} \left(1+\frac{1}{2^n}\right)^{2^n}$$ I already proved it is strictly increasing, but I also have to prove its bounded. I need help with proving it to be bounded.
Prove that $\lim_{n\to \infty}\left (1+\frac{1}{2^n}\right)^{2^n}$ exists
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6Do you recognise some other, famous limit? – 2012-10-23
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0yes, but we cann't use that. I have to prove its monotone and its bounded, then it has a limit – 2012-10-23
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0You could check out [this page](http://www.proofwiki.org/wiki/One_Plus_Reciprocal_to_the_Nth). – 2012-10-23
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0Do you know the inequality $\ln(1+x)\lt x$ for all positive $x$? Your boundedness follows quickly from that. – 2012-10-23
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0@StevenStadnicki how? I don't see how it follows quickly, could you explain? – 2012-10-23
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0@badshah The log of your expression is $2^n \ln (1+\frac{1}{2^n})$; from $\ln(1+x) \lt x$ it then follows that $2^n\ln (1+\frac{1}{2^n})\lt 2^n (\frac{1}{2^n})=1$. Exponentiating both sides of the inequality (using the monotonicity of exp, essentially) gives you a bound. – 2012-10-23
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0Oke, and why is log(1+x)
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1@Badshah For instance, it's true (with $\leq$ rather than $\lt$) when $x=0$, and since $\frac{d}{dx} \ln(1+x) = \frac{1}{1+x} \lt 1 = \frac{d}{dx} x$ then it's true for all larger $x$. – 2012-10-23
3 Answers
Putting $2^n=r$ ,so $r\to \infty$ as $n\to \infty$
So, the limit becomes $\lim_{r\to \infty}(1+\frac 1 r)^r$
Now, its $(s+1)$th term of in binomial expansion is $$\frac{r(r-1)\cdots(r-s)}{r^s(s!)}=\frac 1{s!}\prod_{0\le t\le s}\left(1-\frac t r\right)$$
As $r\to\infty, (s+1)$ th term becomes $\frac 1{s!}$
$$\lim_{r\to \infty}(1+\frac 1 r)^r=\sum_{0\le s< \infty}\frac 1{s!}=1+\frac 1 {1!}+\frac 1 {2!}+\cdots>2$$
Now, $3!=1\cdot2\cdot3>1\cdot2\cdot2\implies \frac 1{3!}<\frac 1{2^2}$
Similarly, $\frac 1{4!}<\frac 1{2^3}, \frac 1{5!}<\frac 1{2^4}$
So, $\sum_{0\le s< \infty}\frac 1{s!}<1+1+\frac 1{2}+\frac 1{2^3}+\frac 1{2^4}+\cdots=1+\frac{1}{1-\frac 1 2}=1+2=3$
So, $$2<\lim_{r\to \infty}(1+\frac 1 r)^r<3$$
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0I cant use that – 2012-10-23
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0@Badshah, could you please share the reason and tell me which formulae are allowed? – 2012-10-23
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0@Badshah, then lookup a proof for this equivalence and apply it to your problem. – 2012-10-23
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0I have to prove its monotone and its bounded. I also know that 1+k/x >=(1+1/x)^k for k a natural number and x>0 – 2012-10-23
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0That's odd. Putting $x = k$ yields $e \le 2$, an apparent contradiction. – 2012-10-23
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0@Badshah the sign in the inequality is reversed. $1 + \frac{k}{x} < (1 + 1/x)^k$ it seems. – 2012-10-23
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0Thanks, this clear and yes, it should be <. – 2012-10-23
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0@Badshah, welcome, did I make any mistake? – 2012-10-23
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0This seems far from being a complete answer by itself; it's missing easily the most important part, that $\lim (1+\frac{1}{r})^r = \sum \frac{1}{s!}$. – 2012-10-23
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0@StevenStadnicki, could you please elaborate a bit. – 2012-10-24
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0@labbhattacharjee Sorry - I should say that that equation is used without any explanation of how it might be derived or where it might come from. _Why_ are the two quantities equal? – 2012-10-24
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0@StevenStadnicki, could you please look into the edited answer? – 2012-10-24
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0@labbhattacharjee That makes it much clearer! +1. – 2012-10-24
To prove it is bounded, we will exploit the fact $ \ln(1+x) \leq x $. Now, we have
$$ a_n = e^{2^n\ln(1+\frac{1}{2^n})} \leq e^{} $$
Let's make two steps together, hopefully that should be enough. Consider $f(x) = (1+1/x)^x$ and it is now sufficient to show that $f(x)$ is bounded as $x \to \infty$. Let $L(x) = \log f(x) = x \log(1+1/x)$. Now since $\log(x)$ is continuous, it suffices to show that $L(x)$ is bounded as $x \to \infty$. (You could e.g. prove that $L(x) < 3$ for all $x > 1$.)