Cesaro Means decrease slower than the sequence

Question

For a strictly decreasing sequence of positive real numbers, I want to show that the Cesaro means decrease slower than the sequence itself. In particular, I need $$\dfrac{C_n}{C_{n+1}}<\dfrac{a_n}{a_{n+1}},$$ where $C_n=\dfrac{1}{n}\displaystyle\sum_{i=1}^na_i.$ This is indeed true for $a_n=\frac{1}{n}$ or $a_n=r^n$ with $r<1$.

Answer 1

This is false. Consider $2,1,1$ and $n=2$. Then $a_2 = a_3 = 1$ while $\frac{C_2}{C_3} = \frac{3/2}{4/3} > 1$.