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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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    We always have $a_n\ge1$, when x>0. Therefore the sequence cannot converge to $x$ unless $x\ge1$.2012-11-12

1 Answers 1

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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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    Don't you see t$h$at it was my question? I repost because noone answered about decreasing or increasing.2012-11-12