Convergence of $(x_n)_n$ if $2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdots x_n}$ $\forall~n$ I became interested in solving a question in a recently deleted post concerning the relation $$2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdot\cdots\cdot x_n}$$ for $n\in\mathbb{N}$. 

The question was whether it converges and if so what the limit is.

My thoughts:
We have $$\begin{array}{c|c}n&x_{n+1}\\\hline1&2018^{2/1}x_1^{-1/2}\\2&2018^{4/3}x_1^{-1/6}\\3&2018^{7/6}x_1^{-1/12}\\...&...\end{array}$$ The pattern is that for $n\in\mathbb{N}$, $$x_{n+1}=2018^{1+[2/n(n-1)]}x_1^{-1/n(n-1)}.$$ Note that we suppose $x_1$ is fixed. This means that as $n\to\infty$, $$x_\infty=2018^{1+0}x_1^{0}=2018$$ so the sequence converges.
Is this correct?
 A: 
Your answer is correct in fact we have the following relations 
   $$\color{blue}{x_{n+2} =2018^{\frac{2}{n+2}}x_{n+1}^{\frac{n}{n+2}}}$$and 
  $$\color{red}{x_{n+1}=2018^{1+\frac{2}{n(n+1)}}x_1^{-\frac{1}{n(n+1)}}.} \to 2018$$
See the prove below

Let us establish a recursive formula3. we have  $$2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdot\cdots\cdot x_n} \implies {x_1\cdot x_2\cdot\cdots\cdot x_n} = \left(\frac{2018^2}{x_{n+1}}\right)^{n+1}$$
then $$x_{n+2} = 2018^2\left(x_1\cdot x_2\cdot\cdots\cdot x_nx_{n+1}\right)^{-\frac{1}{n+2}} = 2018^2\left(\left(\frac{2018^2}{x_{n+1}}\right)^{n+1}x_{n+1}\right)^{-\frac{1}{n+2}} \\= \left(2018^2\right)^{1-\frac{n+1}{n+2}}\left(\left(x_{n+1}\right)^{-n-1}x_{n+1}\right)^{-\frac{1}{n+2}} =2018^{\frac{2}{n+2}}\left(x_{n+1}\right)^{\frac{n}{n+2}} $$
That is we have the recurrence relation, $$\color{blue}{x_{n+2} =2018^{\frac{2}{n+2}}x_{n+1}^{\frac{n}{n+2}}}$$
Now assume that $x_n$ has the form $$x_n =\ell^{a_n}x_1^{b_n}~~~~with~~~~\ell=2018.$$
Obviously $a_2 = 2$ and $b_2= -\frac{1}{2}$
Then by the aforementioned recursive relation we have 
$$\ell^{a_{n+1}}x_1^{b_{n+1}} = x_{n+1}= \ell^{\frac{2}{n+1}}x_{n}^{\frac{n-1}{n+1}}=\ell^{\frac{2}{n+1}}\left(\ell^{a_n}x_1^{b_n}\right)^{\frac{n-1}{n+1}}$$
after identification one remains with
$$a_{n+1} =a_n\frac{n-1}{n+1}+\frac{2}{n+1}~~~and~~~~b_{n+1} =b_n\frac{n-1}{n+1}~~n\ge 2$$
By telescopic product one get, 
$$\color{blue}{\frac{b_{n+1}}{b_2}=\prod^{n}_{k=2}\frac{b_{k+1}}{b_k}=\prod^{n}_{k=2}\frac{k-1}{k+1}=\prod^{n}_{k=2}\frac{k-1}{k}\prod^{n}_{k=2}\frac{k}{k+1} =\frac{2}{n(n+1)}}$$
Hence $$b_{n+1}=- \frac{1}{n(n+1)}$$
One can prove see here that $$a_{n+1}=1+ \frac{2}{n(n+1)}$$
hence
$$\color{red}{x_{n+1}=2018^{1+\frac{2}{n(n+1)}}x_1^{-\frac{1}{n(n+1)}}.} \to 2018$$

Also see this
