limit of a relation

Question

$$ a_2>0,\;a_3>0 ,\; a_{n+2}a_n=(1+1/n)^n a_{n+1}^2, n\ge2 $$

Find the limit of $ \sqrt[n^2]{a_n} $

I used the log and I called $\ln(a_n)$ with $b_n$, then I got $b_{n+2} + b_n = (b_{n+1})^2 $ ...and here I got stucked. Can somebody help me,please?

Answer 1

I believe that the limit is $\sqrt{e}$ by the following handwaving arguments:

The equation can be rewritten as $$ \frac{a_{n+2}}{a_{n+1}} = \left(1 + \frac{1}{n}\right)^n \frac{a_{n+1}}{a_n} $$ or if we set $c_n=a_{n+1}/a_n$, $$ c_{n+1} = \left(1 + \frac{1}{n}\right)^n c_n $$ i.e. $$\frac{c_{n+1}}{c_n} = \left(1 + \frac{1}{n}\right)^n \to e$$

Thus, for big $n$ we will have $c_n \sim A e^n$ for some constant $A$.

Returning back to $a_n$ this means that for big $n$ we have $$ \frac{a_{n+1}}{a_n} \sim Ae^n $$ which implies that $$ a_n \sim Be^{1+2+3+\cdots+n} = B e^{n(n+1)/2} = B e^{n^2/2+n/2} $$ for some constant $B$.

This gives $$ a_n^{1/n^2} \sim B^{1/n^2} e^{1/2+1/(2n)} \to 1 \cdot e^{1/n} = \sqrt{e} $$