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 $a_{n+1}\le\dfrac{1}{n+2}$

We write the recurrence as $$ a_{n+1}=\frac{2n+1}n a_n - \frac1n. $$ Then we divide both sides by $\prod_{j=1}^n \frac{2j+1}j$. We obtain $$ \frac{a_{n+1}}{\prod_{j=1}^n \frac{2j+1}j} = \frac{a_n}{\prod_{j=1}^{n-1}\frac{2j+1}j}-\frac1{(2n+1)\prod_{j=1}^{n-1} \frac{2j+1}j}. $$ Repeatedly applying this, we have $$ \frac{a_{n+1}}{\prod_{j=1}^n\frac{2j+1}j}=\frac{a_2}3-\sum_{k=2}^n \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j} $$ Note that $a_2=3a_1-1$ and $$ \frac13+\sum_{k=2}^{\infty} \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}=2-\frac{\pi}2, $$ as commented on Tesla Daybreak's answer.

Then we need $a_1=2-\frac{\pi}2$ to have a bounded sequence $\{a_n\}$.

$$ \frac{a_{n+1}}{\prod_{j=1}^n\frac{2j+1}j}=2-\frac{\pi}2-(\frac13+\sum_{k=2}^n \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}) $$

The RHS is $$ \sum_{k=n+1}^{\infty} \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j} $$

Thus, $$ a_{n+1}=\sum_{k=n+1}^{\infty} \frac{\prod_{j=1}^n \frac{2j+1}j}{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j} $$

After some cancellations, $$ a_{n+1}=\frac1{2n+3} + \frac1{(2n+5)\frac{2n+3}{n+1}}+\frac1{(2n+7)\frac{2n+5}{n+2}\frac{2n+3}{n+1}}+\cdots. $$ $$ \leq \frac1{2n+3}+\frac1{(2n+5)2}+\frac1{(2n+7)2}\leq \frac1{n+2}. $$

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. $a_n \rightarrow 0$.

This is only a partial solution, but it should be completed without too much extra effort (I currently have not sufficient time to work on it). As pointed out in a comment, it suffices to show $0<a_n\leq\frac1{n+1}$.

First, we let $a_1=t$ be an indeterminate, and write $a_n=x_nt+y_n$ as obtained by the recurrence. Then $$\begin{align}\begin{bmatrix}x_1\\y_1\end{bmatrix}&=\begin{bmatrix}1\\0\end{bmatrix},\\ \begin{bmatrix}x_{n+1}\\y_{n+1}\end{bmatrix}&=\frac{2n+1}n\begin{bmatrix}x_n\\y_n\end{bmatrix}+\begin{bmatrix}0\\-\frac1n\end{bmatrix}.\end{align}$$

Now easily, $$x_n=\prod_{k=1}^{n-1}\frac{2k-1}k=\frac{(2n-1)!}{2^{n-1}[(n-1)!]^2}.$$

Next, define $z_n=-\frac{x_n}{y_n}$. (Keep in mind that we want $x_n+ty_n\rightarrow0$, so $z_n$ should go to our desired initial value.) Substitute $y_n=-x_nz_n$ in the recurrence for $y_n$, and after some calculation, we get $$(z_{n+1}-z_n)x_{n+1}=\frac1n.$$

Hence, $\{z_n\}$ is an increasing sequence. Explicitly, $$z_n=\sum_{k=1}^{n-1}\frac1{kx_{k+1}}=\sum_{k=1}^{n-1}\frac{2^k(k!)^2}{k(2k+1)!}.$$

By comparison test ($\frac1{kx_{k+1}}<\frac1{k^2}$), $\lim_{n\rightarrow\infty}z_n$ exists. Write it as $a$: (We certainly want $a=2-\frac\pi2$.) $$a=\sum_{k=1}^\infty\frac{2^k(k!)^2}{k(2k+1)!}$$ Since the sequence is increasing, $a-z_n>0$ for all $n$.

Now take $a_1=a$ in the original sequence. Then $$a_n=ax_n+y_n=x_n(a-z_n)>0.$$

It remains to show two points: (1) $a=2-\frac\pi2$; (2) $x_n(a-z_n)\leq\frac1{n+1}$.

You can convert the recurrence equation into a differential equation by multiplying it with $x^n$, summing over $n=1,2,3,...$ and defining $f(x)=\sum_{n=0}^\infty a_{n+1} x^n$, which gives the first order DE $$\sum_{n=1}^\infty a_{n+1}x^n = 2\sum_{n=1}^\infty a_n x^n + \sum_{n=1}^\infty \frac{a_n}{n} x^n - \sum_{n=1}^\infty \frac{x^n}{n} \\ \Rightarrow \quad f(x)-a_1=2x f(x) + \sum_{n=1}^\infty \frac{a_n}{n} x^n + \log(1-x) \\ \Rightarrow \quad (1-2x)f'(x) - 3f(x) + \frac{1}{1-x} = 0$$ that can be solved in full generality. You can use that formula or also just check that $$f(x)=-\frac{2\arctan\left(\sqrt{1-2x}\right)}{\left(1-2x\right)^{3/2}} + \frac{2}{1-2x} + \frac{\pi-c}{(1-2x)^{3/2}}$$ solves the ODE where $c$ is an arbitrary constant. The initial condition $$-\frac{\pi}{2}+2 + \pi -c=f(0)=a_1=2-\frac{\pi}{2}$$ implies $c=\pi$ and it is easy to see that this very special case removes the singularity at $x=1/2$. This singularity would lead to the violation of the inequality $a_n \leq \frac{1}{n+1}$ as the coefficients $a_{n+1}$ would then obey $\sim (-2)^n\binom{-3/2}{n}$ asymptotically. Now we express $f(x)$ as an integral and use the geometric series $$\frac{2}{1-2x} - \frac{2\arctan\left(\sqrt{1-2x}\right)}{\left(1-2x\right)^{3/2}} = \int_0^1 \frac{2t^2}{1+(1-2x)t^2} \, {\rm d}t = \sum_{k=0}^\infty x^k \, \underbrace{2^{k+1} \int_0^1 {\rm d}t \, \frac{t^{2k+2}}{(1+t^2)^{k+1}}}_{a_{k+1}}$$

where $0<x<1/2$ was assumed. Interchanging summation and integration is allowed by dom. convergence. Hence it needs to be shown that $$0\leq \frac{1}{n+1} - 2^n \int_0^1 {\rm d}x \, \frac{x^{2n}}{(1+x^2)^{n}} = \int_0^1 {\rm d}x \left( x^n - \frac{2^n x^{2n}}{(1+x^2)^{n}} \right) \\ = \int_0^1 {\rm d}x \, \frac{x^n}{(1+x^2)^n} \left\{ (1+x^2)^n - (1+1)^n x^n \right\} \, .$$ From this it suffices to show

$$0\leq (1+x^2)^n - (1+1)^n x^n \\ = \sum_{k=0}^n \binom{n}{k} \left(x^{2k} - x^n\right) = \sum_{0 \leq k < n/2} \binom{n}{k} (x^{2k}-x^n) + \sum_{n/2 < k \leq n} \binom{n}{k} (x^{2k} - x^n) \\ = \sum_{0 \leq k < n/2} \binom{n}{k} (x^{2k}-x^n) + \sum_{0 \leq k < n/2} \binom{n}{n-k} (x^{2(n-k)} - x^n) =\sum_{0 \leq k < n/2} \binom{n}{k} \left( x^k - x^{n-k} \right)^2 \, .$$