# Prove that $\lim_{n\to \infty}\left (1+\frac{1}{2^n}\right)^{2^n}$ exists

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.

• Do you recognise some other, famous limit? – Lord_Farin Oct 23 '12 at 16:25
• yes, but we cann't use that. I have to prove its monotone and its bounded, then it has a limit – Badshah Oct 23 '12 at 16:26
• You could check out this page. – Lord_Farin Oct 23 '12 at 16:27
• Do you know the inequality $\ln(1+x)\lt x$ for all positive $x$? Your boundedness follows quickly from that. – Steven Stadnicki Oct 23 '12 at 17:26
• @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 $x$. – Steven Stadnicki Oct 23 '12 at 18:05

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, $$2<\lim_{r\to \infty}(1+\frac 1 r)^r<3$$

• I cant use that – Badshah Oct 23 '12 at 16:26
• @Badshah, could you please share the reason and tell me which formulae are allowed? – lab bhattacharjee Oct 23 '12 at 16:28
• @Badshah, then lookup a proof for this equivalence and apply it to your problem. – akkkk Oct 23 '12 at 16:28
• I 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 – Badshah Oct 23 '12 at 16:30
• That's odd. Putting $x = k$ yields $e \le 2$, an apparent contradiction. – Lord_Farin Oct 23 '12 at 16:34

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 $x > 1$.)