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As in the title, I have to prove that $$a_{n}=\sqrt[n]{x^n+x^{n-1}+\ldots+x+1}$$ is decreasing and it goes to $x$. My attempt was to write it as $$a_{n+1}-a_{n}=\sqrt[n+1]{x^{n+1}+a_n^n}-a_{n}$$ however is does not help. I would be very grateful for any suggestion, hints, etc. Thanks in advance!

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    What about $x$ ?2012-11-12
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    Try considering $a_{n+1}/a_n$ instead of $a_{n+1}-a_n.$2012-11-12
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    What assumptions are you making about $x$?2012-11-12
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    $$x^{n+1}-1=(x-1)(x^n+x^{n-1}+ \ldots + 1)$$ $$x^{n}+\ldots + 1= \frac{x^{n+1}-1}{x-1}$$ if $x \neq 1$.2012-11-12
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    $x>0$ and i know sum of geometric serie, but what it gives us?2012-11-12
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    The claim is false. If $x=0$ then $a_n=1$ for all $n$ and converges to $1\ne x$. If $x>0$ then $a_1=1$, $a_2=\sqrt{1+x}>a_1$, hence the sequence is not decreasing.2012-11-12
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    Could be. So you suggest using induction to prove that it increases?2012-11-12
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    We always have $a_n\ge1$, when $x>0$. Therefore the sequence cannot converge to $x$ unless $x\ge1$.2012-11-12

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For $x>1$, observe that $$b_n:=\frac{a_n}x=\sqrt[n]{1+\frac1x+\cdots +\frac1{x^n}}$$ and the radicand converges to $\frac1{1-\frac1x}=\frac x{x-1}>0$, hence $b_n\to 1$ and $a_n\to x$.

If $x=1$ then $a_n=\sqrt[n] n\to1$ is well-known.

If $-1, then $$a_n=\sqrt[n]{1+x+\cdots + x^n}=\sqrt[n]{\frac{1-x^{n+1}}{1-x}}\to 1$$ because $\sqrt[n]{1-x}\to 1$ and even more so $\sqrt[n]{1-x^{n+1}}\to 1$ because $1-x^{n+1}\to1$.

In summary:

  • $a_n\to x$ if $x\ge 1$
  • $a_n\to 1$ if $-1< x\le 1$.
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    Yeah, but the task was to prove that is decreases/increases, not to find its limits (however I appreciate that you did this, thanks for it)2012-11-12
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    @user46034 One of the tasks was to show "and it goes to $x$" (which it doesn't always do). The other task is to show that it decreases (which it doesn't always do either).2012-11-12
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    @Sergei $\lim a_n=1$ if $x=0$.2012-11-12
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    Alright, my mistake (you are right). But how to prove decreasing (or increasing) in such a case?2012-11-12
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    Same question was asked recently, [please see this.](http://math.stackexchange.com/questions/234821/examine-sqrtnxnxn-1-x1)2012-11-12
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    @HagenvonEitzen yep2012-11-12
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    Don't you see that it was my question? I repost because noone answered about decreasing or increasing.2012-11-12