# Cesaro Means decrease slower than the sequence

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 $$r<1$$.

## 1 Answer

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

• I meant strictly decreasing. Sorry for the confusion, will edit. – Arnab Auddy Nov 29 '18 at 18:00
• It's still false, by continuity. Just change the sequence to $2,1,.9999$ – mathworker21 Nov 29 '18 at 18:00
• You are correct, thanks. Where do I look for a sufficient condition? I need it for sequences like $a_n=log(A/n)^2$ with fixed A. – Arnab Auddy Nov 29 '18 at 18:20
• I think you want $\sum_{n \le N} a_n \le \frac{Na_Na_{N+1}}{N(a_{n+1}-a_n)+a_{n+1}}$ if the denominator is positive. If the denominator is negative, then the original inequality is true. So it looks false for $\log(A/n)^2$. – mathworker21 Nov 29 '18 at 18:34