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!
$a_{n}=\sqrt[n]{x^n+x^{n-1}+\ldots +x+1}$ prove decreasing
2
$\begingroup$
real-analysis
limits
-
0What about $x$ ? – 2012-11-12
-
0Try considering $a_{n+1}/a_n$ instead of $a_{n+1}-a_n.$ – 2012-11-12
-
0What assumptions are you making about $x$? – 2012-11-12
-
0$$x^{n+1}-1=(x-1)(x^n+x^{n-1}+ \ldots + 1)$$ $$x^{n}+\ldots + 1= \frac{x^{n+1}-1}{x-1}$$ if $x \neq 1$. – 2012-11-12
-
0$x>0$ and i know sum of geometric serie, but what it gives us? – 2012-11-12
-
0The 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
-
0Could be. So you suggest using induction to prove that it increases? – 2012-11-12
-
0We always have $a_n\ge1$, when $x>0$. Therefore the sequence cannot converge to $x$ unless $x\ge1$. – 2012-11-12
1 Answers
1
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
In summary:
- $a_n\to x$ if $x\ge 1$
- $a_n\to 1$ if $-1< x\le 1$.
-
0Yeah, 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
-
0@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
-
0@Sergei $\lim a_n=1$ if $x=0$. – 2012-11-12
-
0Alright, my mistake (you are right). But how to prove decreasing (or increasing) in such a case? – 2012-11-12
-
0Same question was asked recently, [please see this.](http://math.stackexchange.com/questions/234821/examine-sqrtnxnxn-1-x1) – 2012-11-12
-
0@HagenvonEitzen yep – 2012-11-12
-
0Don't you see that it was my question? I repost because noone answered about decreasing or increasing. – 2012-11-12