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