Sequences

Consider a_1,\cdots ,a_n,a_{n+1}\in \left \{ a_r \right \}.
Let (\alpha_1,\alpha_2,\cdots ,\alpha_n) be a permutation of (a_1,a_2,\cdots,a_n), such that \alpha_1\leq \alpha_2\leq \cdots \leq a_n, it still follows that (a_{t+1}- a_t)\leq 1, \forall t\in \left \{ 1,n-1 \right \}. Also (\alpha_1-1)\leq a_{n+1}\leq (\alpha_n+1 ).

Note that (b_{n}- b_{n+1}) = \frac{\left ( \sum_{i=1}^{n}\alpha_i)- n\cdot a_{n+1}\right )}{n(n+1)}

We have \frac{\left ( \sum_{i=1}^{n}\alpha_i)- n\cdot a_{n+1}\right )}{n(n+1)}\leq \frac{\left( \sum_{r=0}^{n-1}(\alpha_1+ r)- n(\alpha_1-1)\right )}{n(n+1)}= \frac{1}{2}

On the other hand \frac{\left ( \sum_{i=1}^{n}\alpha_i)- n\cdot a_{n+1}\right )}{n(n+1)}\geq \frac{\left ( \sum_{r=0}^{n-1}(\alpha_n- r)- n\cdot (\alpha_n + 1) \right )}{n(n+1)}= \frac{-1}{2}

So we have proved that |b_n- b_{n+1}| \le \frac{1}{2}

another approach,
given that |a_n-a_n+1|≤ 1
|a₁-a₂|≤1
|a₂-a₃|≤1
.
.
.
|a_n-a_n+1|≤1
on summing up and using the triangle inequality,
|a₁-a_n+1| ≤n ; |a₂-a_n+1|≤(n-1) similarly others ....(*)
now on arranging,
|b_n-b_n+1|
= |(a₁-a_n+1) + (a₂-a_n+1).....(a_n-a_n+1)|n(n+1)

= again inequality
≤ 1n(n+1){|a₁-a_n+1| +|a₂-a_n+1|......} 1
≤ on using (*)
≤ 1n(n+1) * Σ(n-r) (from r=0 -> r=n-1)
≤ 1n(n+1)*n(n+1)2
≤12
Q.E.D

\left|b_{n+2}-b_{n+1} \right|

=\left|\frac{a_{1}+a_{2}+....+a_{n+1}+a_{n+2}}{n+2}-b_{n+1} \right|
=\left|\frac{a_{1}+a_{2}+....+a_{n+1}}{n+1}\times\frac{n+1}{n+2}+\frac{a_{n+2}}{n+2}-b_{n+1} \right|
=\left|b_{n+1}\times\frac{n+1}{n+2}-b_{n+1}+\frac{a_{n+2}}{n+2} \right|
=\left|\frac{a_{n+2}}{n+2}-\frac{b_{n+1}}{n+2} \right|
=\left|\frac{a_{n+2}-\frac{a_{1}+a_{2}+....+a_{n+1}}{n+1}}{n+2} \right|
=\left|\frac{(a_{n+2}-a_{n+1})+(a_{n+2}-a_{n})+(a_{n+2}-a_{n-1})+....+(a_{n+2}-a_{2})+(a_{n+2}-a_{1})}{(n+1)(n+2)} \right|
=\left| \frac{[(a_{n+2}-a_{n+1})]+[(a_{n+1}-a_{n})+1]+[(a_{n}-a_{n-1})+2]+....+[(a_{2}-a_{1})+n]}{n+2}\right|
\leq\left|\frac{ [ 1]+[1+1]+[1+2]+....+[1+n]}{n+2} \right| [|a_n-a_n+1|≤1|]
=\left|\frac{(1)+(2)+....+(n)+(n+1)}{(n+1)(n+2)} \right|
=\left|\frac{\frac{(n+1)(n+2)}{2}}{(n+1)(n+2} \right|
=\frac{1}{2}