<s>[INST] show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$ let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{3}{2}\pi$$
so
$$f(2)=\lceil \dfrac{2}{\dfrac{3\pi}{2}-4}\rceil=3$$
and so on ,if we want prove
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
or
$$n<\dfrac{n}{1-a_{n}}\le n+1$$
or
$$0<a_{n}\le\dfrac{1}{n+1}$$
if we use induction to prove it.then
$$a_{n+1}=(2+\dfrac{1}{n})a_{n}-\dfrac{1}{n}\le (2+\dfrac{1}{n})\cdot\dfrac{1}{n+1}-\dfrac{1}{n}=\dfrac{1}{n+1}$$
is wrong,becasue we want to prove $a_{n+1}\le\dfrac{1}{n+2}$
 [/INST] The series $\frac{\pi}{2}=\frac{1}{2}\sum\limits_{n=0}^\infty \frac{(n!)^22^{n+1}}{(2n+1)!}=1+\frac{1}{3}+\frac{1\cdot 2}{3\cdot 5}+\frac{1\cdot2\cdot3}{3\cdot5\cdot7}+\cdot\cdot\cdot$
causes $2-\frac{\pi}{2}$ to have nice properties with the given sequence $a_{n+1}= \frac{2n+1}{n}a_n-\frac{1}{n}.$
$a_1 = 2-\frac{\pi}{2}=1-\frac{1}{3}-\frac{1\cdot 2}{3\cdot 5}-\frac{1\cdot2\cdot3}{3\cdot5\cdot7}-\cdot\cdot\cdot$
$a_2=1-\frac{2}{5}-\frac{2\cdot 3}{5\cdot7}-\cdot \cdot \cdot$
$\vdots$
$a_n=1-\frac{n}{2n+1}-\frac{n(n+1)}{(2n+1)(2n+3)}-\cdot \cdot \cdot$
from which $a_n >1-\frac{1}{2}-\frac{1}{2^2}-\cdot \cdot \cdot = 1-\sum\limits_{i=1}^\infty 2^{-i}=1-1=0.$
Note that if $a_n$ is a decreasing sequence implies $a_n \leq \frac{1}{n+1}$
$a_{n+1}\leq a_n \Longleftrightarrow$
$\frac{2n+1}{n}a_n -\frac{1}{n} \leq a_n \Longleftrightarrow$
$(2n+1)a_n-1\leq na_n \Longleftrightarrow$
$(n+1)a_n\leq 1 \Longleftrightarrow$
$a_n\leq \frac{1}{n+1}$
and the first inequality
$a_{n+1}=1-\frac{n+1}{2n+3}-\frac{(n+1)(n+2)}{(2n+3)(2n+5)}-\cdot \cdot \cdot \leq a_n =1-\frac{n}{2n+1}- \frac{n(n+1)}{(2n+1)(2n+3)}-\cdot \cdot \cdot$ does hold because for each term
$\frac{(n)(n+1)(n+2)\cdot\cdot\cdot(n+m)}{(2n+1)(2n+3)(2n+5)\cdot \cdot \cdot (2n+2m+1)}\leq \frac{(n+1)(n+2)\cdot\cdot\cdot(n+m)(n+m+1)}{(2n+3)(2n+5)\cdot \cdot \cdot (2n+2m+1)(2n+2m+3)}$
and cancelling like terms it is true that
$\frac{n}{2n+1} \leq \frac{n+m+1}{2n+2m+3}$ for $m\geq0.$
Although induction is tempting, I think all the properties of the sequence are captured in the series $\frac{\pi}{2}=\frac{1}{2}\sum\limits_{n=0}^\infty \frac{(n!)^22^{n+1}}{(2n+1)!}$.
Note that I proved also that $a_n$ is bounded below and decreasing, and thus convergent. Instead of looking at the inequalities $0<a_n\leq \frac{1}{n+1}$ as a possible induction proof, you can look at them as squeeze inequalities, i.e. $a_n \rightarrow 0$.
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