What's the limit of the string?

Question

Find $$\lim_{n\to\infty} (x_n\sqrt{n})^{\sqrt{n^2-1}},$$ where $ x_{n+1} = \frac{x_n}{\sqrt{1+x_n^2}}$ and $x_1 = 2$.

I showed $x_n \to 0$, $x_n\sqrt{n} \to 1$, but i don't know how to solve limit properly.

Answer 1

Later EDIT: In the initial post there was an error, the old computation was using $y_0=2$, instead of $y_{\color{red}1}={\color{red}4}$.

Let the sequence $(y_n)$ be given by $y_n=x_n^2$. Then we have the Möbius transformation action giving the recursion: $$ y_n =\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}\cdot y_{n-1}\ . $$ Here, the action of the $2\times 2$-matrix with entries $a,b,c,d$ on an element in $z\in \Bbb R$ (or $\Bbb C$) is given by $$ \begin{bmatrix} a & b\\ c&d \end{bmatrix}\cdot z =\frac {az+b}{cz+d}\ . $$ This gives $$ \begin{aligned} y_n &= \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}^{n-1}\cdot y_1 = \begin{bmatrix} 1 & 0\\ n-1 & 1 \end{bmatrix}\cdot 4 \\ &= \frac {4}{4n-3}\text{ . So we have:} \\ x_n &= \sqrt{\frac {4}{4n-3}}\ . \end{aligned} $$ (In case the argument using Möbius transformations is to heavy, just use induction to show the above formula.)

So we have to compute the limit of $$ (x_n\sqrt{n})^{\sqrt{n^2-1}} = \left(\sqrt{\frac {4n}{4n-3}}\right)^{\sqrt{n^2-1}} = \left(1+\frac 3{4n-3}\right)^{\frac 12\sqrt{n^2-1}} = \left(1+\frac 3{4n-3}\right)^{(4n-3)\frac 1{2(4n-3)}\sqrt{n^2-1}} = \left[\ \left(1+\frac 3{4n-3}\right)^{(4n-3)}\ \right]^{\frac 1{2(4n-3)}\sqrt{n^2-1}} \ . $$ The limit is now easy to compute, the $[\dots]$ expression goes to $e^3$, and its exponent to $\frac 18$, so
the limit is $\exp\frac 38$.

Answer 2

Strating from @dan_fulea nice solution $$x_n= \sqrt{\frac {4}{4n-3}}$$we could have nice approximation of $$y_n=\big(x_n\sqrt{n}\big)^{\sqrt{n^2-1}}=\left(1+\frac 3{4n-3}\right)^{\frac 12\sqrt{n^2-1}}$$ Take logarithms $$\log(y_n)={\frac 12\sqrt{n^2-1}}\,\log\left(1+\frac 3{4n-3}\right)$$

Now, use Taylor series for large values of $n$ $$\sqrt{n^2-1}=n-\frac{1}{2 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$ $$\log\left(1+\frac 3{4n-3}\right)=\frac{3}{4 n}+\frac{9}{32 n^2}+\frac{9}{64 n^3}+O\left(\frac{1}{n^4}\right)$$ $$\log(y_n)=\frac{3}{8}+\frac{9}{64 n}-\frac{15}{128 n^2}+O\left(\frac{1}{n^3}\right)$$ $$y_n=e^{\log(y_n)}=e^{3/8}\left(1+\frac{9}{64 n}-\frac{879}{8192 n^2}+O\left(\frac{1}{n^3}\right) \right)$$

Just for the fun, using your pocket calculator for $n=10$; the exact value is $1.47381$ whle le above approximation gives $1.47389$.