Show sequance is monotonic Let $x>0$ (fixed) and $n$ be natural. Show that $$\displaystyle (x^n+x^{n-1}+...+1)^{\frac{1}{n}}$$ is monotonic. 
I tried by induction but didn't work but intuition tells me it's decreasing. 
 A: use that $\sum_{i=0}^n x^i=\frac{x^{n+1}-1}{x-1}$ for $x\ne 1$
A: For $x=1$, $\sqrt[n]{n}$ is incsreasing since $\frac{d}{{dy}}\left( {\sqrt[y]{y}} \right) = \sqrt[y]{y}\frac{{1 - \ln y}}{{{y^2}}} \leqslant 0$ for $e \geqslant y \geqslant 1$ and is decreasing for $y > e$. Indeed, the first four members of the sequence are 1, 1.41421, 1.44225 and 1.41421, so the sequence is not monotone, although it is monotone after a first couple of members.
If not, then ${a_n} = {\left( {\frac{{{x^{n + 1}} - 1}}{{x - 1}}} \right)^{\frac{1}{n}}}$, so 
$\frac{d}{{dy}}{\left( {\frac{{{x^{y + 1}} - 1}}{{x - 1}}} \right)^{\frac{1}{y}}} = \underbrace {{{\left( {\frac{{{x^{y + 1}} - 1}}{{x - 1}}} \right)}^{\frac{1}{y}}}\frac{1}{{{y^2}}}}_{ \geqslant 0}\left( {\frac{{\ln {x^y}}}{{{x^{y + 1}} - 1}} + \ln \left( {x - 1} \right) + \ln \frac{{{x^y}}}{{{x^{y + 1}} - 1}}} \right)$
Similarly as before we obtain, for $y$ sufficiently large, 
$0 < x < 1 \Rightarrow \underbrace {y\left( {1 - \frac{1}{{1 - {x^{y + 1}}}}} \right)\ln x}_{ \leqslant 0} - \underbrace {\ln \left( {1 - {x^{y + 1}}} \right)}_{ \to 0} + \underbrace {\ln \left( {1 - x} \right)}_{ \leqslant 0}$
and
$x > 1 \Rightarrow \frac{{\ln {x^y}}}{{{x^{y + 1}} - 1}} + \ln \frac{{{x^{y + 1}} - x}}{{{x^{y + 1}} - 1}} \geqslant \frac{{\ln {x^y}}}{{{x^{y + 1}} - 1}} + \ln \frac{{{x^{y + 1}} - 1}}{{{x^{y + 1}} - 1}} = \frac{{y\ln x}}{{{x^{y + 1}} - 1}} \geqslant 0$.
So, $\frac{d}{{dy}}{\left( {\frac{{{x^{y + 1}} - 1}}{{x - 1}}} \right)^{\frac{1}{y}}}$ is either negative for sufficiently large $y$, or positive, so that $f\left( y \right) = {\left( {\frac{{{x^{y + 1}} - 1}}{{x - 1}}} \right)^{\frac{1}{y}}}$ is either eventually descending or ascending. Since the sequence in question is the restriction ${\left. f \right|_\mathbb{N}}$, we conclude that the sequence is eventually decreasing for $0 < x < 1$ and always increasing for $x > 1$.
A: The first couple are decreasing at least:
$(x+1)^2-(x^2+x+1)^1=x$
$(x^2+x+1)^3-(x^3+x^2+x+1)^2=x^5+3x^4+3x^3+3x^2+x\\=x(x+1)(x^3+x^2+x+1)+x^2(x^2+x+1)$
Next is $(x^3+x^2+x+1)^4-(x^4+x^3+x^2+x+1)^3$, which is
$$x^{11}+4x^{10}+10x^9+16x^8+22x^7+25x^6+22x^5+16x^4+10x^3+4x^2+x$$
