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judgeAs written above, I have to examine sequence $$a_{n}=\sqrt[n]{x^n+x^{n-1}+...+x+1}$$ I think that it converges to x (substituting) and serie is decreasing. I tried to prove it by writing it as sum of geometric serie, so that $a_{n}=\sqrt[n]{\frac {x^{n+1}-1}{x-1}}$ . Then I defined $b_{n}=ln(a_{n})\Rightarrow b_{n}=\frac{1}{n}ln(\frac{x^{n+1}-1}{x-1})=\frac{1}{n}\left ( ln(x^{n+1}-1)-ln(x-1) \right )$
I am not allowed to use Hospital's rule (it would not change much) and I do not know how to evaluate is further.

Second attempt was based on means: Power mean seems to be greater that $a_{n}$ which is greater than arithmetic mean. But the difference is getting larger and larger instead of useful smaller and smaller.

I also tried something similiar to triangle inequality:
$\sqrt[n]{x^n}\leq\sqrt[n]{x^n+x^{n-1}+...+x+1} \leq \sqrt[n]{x^n} +\sqrt[n]{x^{n-1}}+...+\sqrt[n]{1}$
Here both right and left sides goes to (I hope at least that they do) $x$. I did not go any further, can you judge the (in)correctness of my trials?

Second part of question was to examine if it decreases or increases. And $a_{n+1}-a_{n}=\sqrt[n+1]{x^{n+1} +a_{n}^{n}}-a_{n}$ didn't help what makes me a little helpless. Can somebody help me? Thanks in advance for your time and hints.

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2 Answers 2

up vote 7 down vote accepted

If you assume that $x\geq 1$, then

$$x=\sqrt[n]{x^n}\leq \sqrt[n]{x^n+x^{n-1}+\ldots +x +1} \leq \sqrt[n]{(n+1)x^n}=x\sqrt[n]{n+1}$$

and so by the squeeze theorem, the limit approaches x. If $0<x<1$, then $1<a_n<\sqrt[n]{1/(1-x)}$, and the squeeze theorem gives that the sequence approaches $1$.

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Yes, but i realised that i am still not able to prove that it is decreasing. Limit is clearly explained, thank you both. I tried to write it as $a_{n+1}-a_{n}=\sqrt[n+1]{x^{n+1} +a_{n}^{n}}-a_{n}$, what however was not very useful. Any ideas? Thanks in advance –  fdhd Nov 12 '12 at 15:03
Here are three ideas which I haven't worked out, but which maybe you haven't tried yet: (1) Since they are positive, you can just look at the ratio. (2) Since $x^n$ is a monotonic function, you could look at something like $a_n^n-a_{n-1}^n$. (3) Using the formula for $a_n$ based on the sum of a geometric series, you can treat $x$ as a constant and take the derivative with respect to $n$. It may help to take the natural log first and simplify before taking the derivative. –  Aaron Nov 12 '12 at 21:47

$$\root n\of{x^{n+1}-1\over x-1}=\root n\of{x^{n+1}-1}/\root n\of{x-1}$$ I assume you can handle $\root n\of{x-1}$. And you can compare $\root n\of{x^{n+1}-1}$ with $\root n\of{x^{n+1}}$.

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