I am looking for Arithmetic mean - Harmonic mean - geometric mean and root mean square for the numbers in $(0,1)$. Am I doing it right?
As the first step, I take a partition for the numbers $\{\frac 1n ,\frac 2n ,\frac3n ,\cdots,\frac nn \}$ (as $n$ tens to $\infty) $
Arithmetic mean $$AM=\frac{\Sigma x_i}{n}=\frac{\frac 1n +\frac 2n +\frac3n +\cdots+\frac nn}{n}\\=\frac{\frac{n(n+1)}{2n}}{n}\ \to \frac12$$
Harmonic mean $$HM=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\cdots +\frac{1}{x_n}}\\=
\frac{n}{\frac{1}{\frac 1n}+\frac{1}{\frac 2n}+\frac{1}{\frac 3n}+\cdots +\frac{1}{\frac nn}}\\=
\frac{1}{\frac 11 +\frac12 +\frac13 +\cdots +..\frac 1n}\\ \text{tends to } \frac{1}{\infty}\to0$$
Geometric mean $$GM=\sqrt[n]{\frac 1n .\frac 2n .\frac3n .\cdots.\frac nn }\\ \ln(GM)=\frac 1n (\ln(\frac1n)+\ln(\frac {2}n)+\ln(\frac3n)+\cdots +\ln(\frac nn))\\=
\frac 1n (\Sigma_{i=1}^{n}\ln(\frac in)) \\ \text{rieman sum and tends to } \to \ln(\frac 1e) \\ \to GM=\frac 1e$$
Root mean square $$RMS=\sqrt{\frac {x_1^2+x_2^2+\cdots+x_n^2}{n}}\\=\sqrt{\frac {(\frac1n)^2+(\frac2n)^2+\cdots+(\frac nn)^2}{n}} \\=\sqrt{\frac {\frac{n(n+1)(2n+1)}{6}}{n^3}}\\ \to \sqrt \frac{1}{3}$$