Convergence of $(x_n)_n$ if $2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdots x_n}$ $\forall~n$

Question

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?

@GuyFsone The person wanted to know whether $x_n$ converges as $n\to\infty$. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jan 29 '18 at 19:35
It doesn't have to be a recursive formula, we can still ask whether $x_n$ convergence... — Yanko, Jan 29 '18 at 19:35
But $x_n$ isn't defined, so the question doesn't make any sense. — saulspatz, Jan 29 '18 at 19:36
@TheSimpliFire Do we know that $x_i>0$ for every $i$? otherwise there are problems with the square. — Yanko, Jan 29 '18 at 19:36
@saulspatz It was supposed that the first term $x_1$ was fixed (ie. a constant). — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jan 29 '18 at 19:36
@saulspatz $x_n$ is a sequence satisfying the equality, there could be one, there could be more... — Yanko, Jan 29 '18 at 19:36
@TheSimpliFire OK. The formula is the defining equation of the $x_n$. I'm with you now. — saulspatz, Jan 29 '18 at 19:39
Here is an answer to your question (since I see none on the page at present, although you already accepted a post below, which is odd): If you proved the formula after "The pattern is that" (by opposition to, say, only observing that it works for the first values), for example using an induction on $n$, then indeed you have a proof that $x_n\to2018$ for every initial condition $x_1>0$. So... did you actually proved this explicit formula, by induction or otherwise? — Did, Jan 29 '18 at 20:19
@GuyFsone Not a counterexample, please see my comment to your post. — Did, Jan 29 '18 at 20:43
@martycohen I don't know, I didn't make the question up. But most possibly it is currently the year 2018... — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jan 31 '18 at 18:44

Guy Fsone · Accepted Answer · 2018-01-31T19:18:17.100

3

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

edited Jan 31 '18 at 19:18

answered Jan 29 '18 at 19:39

Guy Fsone

23,903

(+1) It is nice that the link sends to one of your old question. – Yanko Jan 29 '18 at 19:44
@yanko you are welcome It is surpinsing to see only one vote from two differents voters:) – Guy Fsone Jan 29 '18 at 19:51
@The SimpliFire Yes, your method is correct. – saulspatz Jan 29 '18 at 19:54
This does not prove the convergence, does it? – Did Jan 29 '18 at 19:56
This prove that if the limit exists, it must be 2018, but I don't see where it proves existence. – saulspatz Jan 29 '18 at 19:57
@saulspatz you are but see the edit. the converse of this is not true just take
$$x_{2n }= 2~~~~and ~~~x_{2n+1 } =\frac{1}{2}$$

we have $$x_{n+1}^{1-\frac{1}{n+1}}\sqrt[n+1]{x_1\cdot x_2\cdot\cdots\cdot x_nx_{n+1}} \to 1$$

but $\lim_{n\to\infty}x_n$ does not exists.
– Guy Fsone Jan 29 '18 at 20:28
@Did see the above comment and the edit – Guy Fsone Jan 29 '18 at 20:30
@TheSimpliFire Now I do have an objection since there is a counter example – Guy Fsone Jan 29 '18 at 20:31
How is this a counterexample? This sequence does not satisfy the identities $2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdot\cdots\cdot x_n}$ for every $n$ positive, does it? – Did Jan 29 '18 at 20:42
@Did you where right for this case it hold true I have changed the post thanks very much – Guy Fsone Jan 29 '18 at 21:14

Convergence of $(x_n)_n$ if $2018^2=x_{n+1}\sqrt[n+1]{x_1\cdot x_2\cdots x_n}$ $\forall~n$

1 Answers1

See the prove below