An expression for $\lim_{n\to\infty}\frac1{2^n}\left(1 + x^{1/n}\right)^n$

Question

I am looking for a closed form answer to the limit ($x<1$): $$\lim_{n\to\infty}\frac1{2^n}\left(1 + x^{1/n}\right)^n$$

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For context, I was studying weighted averages and considered $$\left(\frac12(x^{1/n} + y^{1/n})\right)^{n}$$ to be a good way to weight averages in favour of the lower number (similar to root mean squared, but kinda reversed). I think was studying what happens for different values of $n$. For $x=3$, $y=5$, I found that this limit seems to converge to $3.78962712197\dots$ but I do not recognise where this number comes from. Rearranging the above average formula gives $$\frac{y}{2^n}\left(1+\left(\frac xy\right)^{1/n}\right)^n$$and this is what inspired the question. I see it looks similar to some kind of exponential, but it isn't quite there. I also tried exponentiating and logging the whole expression to bring down the $n$, but I didn't know how to deal with the power inside the brackets then. My main issue is that since $1/n$ goes to $0$, the thing thats raised to this power goes to $1$ (and so is not small) so series expansions can't be used.

Thanks!

Answer 1

We have for $x>0$ that $$ \frac{1}{{2^n }}\left( {1 + x^{1/n} } \right)^n = \left( {1 + \frac{{x^{1/n} - 1}}{2}} \right)^n = \left( {1 + \frac{{1 + \frac{1}{n}\log x + O\left( {\frac{{\log ^2 x}}{{n^2 }}} \right) - 1}}{2}} \right)^n = \left( {1 + \frac{{\log \sqrt x }}{n} + O\left( {\frac{{\log ^2 x}}{{n^2 }}} \right)} \right)^n \to e^{\log \sqrt x } = \sqrt x . $$

Answer 2

Let $$a_n=2^{-n} \left(1+x^{\frac{1}{n}}\right)^n\implies \log(a_n)=-n\log(2)+n\log \left(1+x^{\frac{1}{n}}\right)$$ Now, use Taylor expansion for large $n$ to get $$\log(a_n)=\frac{\log (x)}{2}+\frac{\log ^2(x)}{8 n}+O\left(\frac{1}{n^3}\right)$$

So, the limit is $\sqrt x$.

Edit

For your curiosity, $$\log(a_n)=\frac 12\sum_{k=0}^\infty \frac{E_k(1) \log ^{k+1}(x)}{ (k+1)! \,n^k}$$ where appear Euler polynomials.

Answer 3

Since the logarithm is continuous, (Interchanging limits and logarithms), with $f=\frac{1}{2^n}(1+x^{1/n})^n$

$$\log \lim_{n\rightarrow \infty} f = \lim_{n\rightarrow \infty}\log f$$

$$\log f = \frac{\log (1+x^{1/n}) - \log 2}{\frac{1}{n}} $$

Applying L'Hopital's rule:

$$\log f \rightarrow \lim_{n\rightarrow \infty}\frac{x^{1/n-1}}{1+x^{1/n}}\log x=\frac{1}{2} \log x $$

Thus

$$f \rightarrow x^{1/2}$$

Answer 4

This is a bit of a hand-wavy argument but I thought it was fun so I'll share it. Let $f_n(x)=\dfrac{\left(1+x^{\frac{1}{n}}\right)^{n}}{2^{n}}$, and assume $\lim\limits_{n\to\infty}f_n(x)=f(x)$ exists, and that the limit of derivatives, $\lim\limits_{n\to\infty}f_n'(x)$, exists and is equal to $f'(x)$. Then, $$\begin{align}f_n'(x)&=\underbrace{\frac{(1+x^{\frac{1}{n}})^n}{2^n}}_{f_n(x)}\cdot\frac{x^{\frac{1}{n}}}{x\left(1+x^{\frac{1}{n}}\right)} \\ \lim_{n\to\infty}f_n'(x)&=\lim_{n\to\infty}f_n(x)\cdot \lim_{n\to\infty}\frac{x^{\frac{1}{n}}}{x\left(1+x^{\frac{1}{n}}\right)} \\ f'(x)&=\frac{f(x)}{2x} \\ \int \frac{f'(x)}{f(x)}\ \mathrm{d}x&=\int\frac{1}{2x}\ \mathrm{d}x \\ \ln |f(x)|&=\frac{1}{2}\ln |x|+C \\ f(x)&=\pm C_2\sqrt{\left|x\right|} \end{align}$$

Then, given $f(1)=\lim\limits_{n\to\infty}f_n(1)=\lim\limits_{n\to\infty} 1$ and that $f,x\ge 0$, we see $f(x)=\sqrt{x}$.