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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 $r<1$.

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

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  • $\begingroup$ I meant strictly decreasing. Sorry for the confusion, will edit. $\endgroup$ – Arnab Auddy Nov 29 '18 at 18:00
  • $\begingroup$ It's still false, by continuity. Just change the sequence to $2,1,.9999$ $\endgroup$ – mathworker21 Nov 29 '18 at 18:00
  • $\begingroup$ 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. $\endgroup$ – Arnab Auddy Nov 29 '18 at 18:20
  • $\begingroup$ 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 $\log(A/n)^2$. $\endgroup$ – mathworker21 Nov 29 '18 at 18:34

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