# Inequality for term of a positive sequence : show $\frac{1}{n} \ge c_n - c_{n+1} \ge \frac{1}{n+1}$

Let $\{x_n\}$ be a sequence in $\mathbb{R}_+^\mathbb{Z}$. Define
\begin{align} a_n&=\sup_{k \ge n} \{x_k\} \\ b_n&=a_n+\frac{1}{n} \\ c_1&=b_1 \text{ and } c_n=\max\left(c_{n-1}-\frac{1}{n},b_n\right) \end{align} Show $$\frac{1}{n} \ge c_n - c_{n+1} \ge \frac{1}{n+1}$$

Writing \begin{align} c_n &\ge b_{n+1}, \\ c_n &\ge c_{n-1}-\frac{1}{n} \end{align} I get $$c_n-c_{n+1} \le c_n - c_ n+ \frac{1}{n+1} = \frac{1}{n+1} \le \frac{1}{n}$$ I don't see how to get the lower bound.

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there is a typo in the definition of c_n –  saposcat Oct 15 '12 at 13:09
and also, i guess a_n=sup_{k\geq n}x_k –  saposcat Oct 15 '12 at 13:10
@sasposcat Good eyes, thanks. –  Nicolas Essis-Breton Oct 15 '12 at 13:27
A lower bound $1/(n+1)$ is most unlikely since you proved that $1/(n+1)$ is an upper bound of $c_n-c_{n+1}$. –  Did Oct 15 '12 at 13:41
The stated lower bound can fail. Take $x_n=1$ for all $n\in\Bbb N$. Then for $n\in\Bbb N$ we have $a_n=1$ and $b_n=1+\frac1n$. Finally, $c_1=b_1=2$, $$c_2=\max\left\{2-\frac12,1+\frac12\right\}=\frac32\;,$$ and $$c_3=\max\left\{\frac32-\frac13,1+\frac13\right\}=\max\left\{\frac76,\frac43\right\}=\frac43\;,$$ so
$$c_2-c_3=\frac32-\frac43=\frac16<\frac13\;.$$