Prove Convergence or divergence of $a_{n+1}=\frac{2n+1}{n}a_{n}-1$

Question

Prove Convergence or divergence of $$a_{n+1}=\frac{2n+1}{n}a_{n}-1$$

I thought $$\lim_{n \to \infty}\frac{2n+1}{n}=2$$ So, if $$\frac{1}{2}{\approx}a_n>\frac{1}{3}$$ Because $$1<\frac{2n+1}{n}a_{n}<2$$ Than $a_n$ is converge. But I can't move forward after this...

Answer 1

Incomplete answer ...

---

From $$a_{n+1}=\frac{2n+1}{n}a_{n}-1\iff a_{n+1}-a_n=\frac{n+1}{n}a_n-1$$ we have $$a_{n+1}-a_1=\sum\limits_{i=1}^n\left(\frac{i+1}{i}a_i-1\right)\Rightarrow \lim\limits_{n\to\infty}a_{n+1}=a_1+\sum\limits_{i=1}^\infty\left(\frac{i+1}{i}a_i-1\right) \tag{1}$$

---

Proposition 1. If $\exists n_0$ s.t. $a_{n_0}\geq 1$ then $a_n\geq 1,\forall n\geq n_0$.

Results from the induction $$a_{n+1}=\frac{2n+1}{n}a_{n}-1 \geq \frac{2n+1}{n}-1=\frac{n+1}{n}>1$$

---

Proposition 2. If $\exists n_0$ s.t. $a_{n_0}\geq 1$ then $(a_n)_{n\in\mathbb{N}}$ is divergent.

From $(1)$, $\forall n \geq n_0$ $$\color{red}{a_{n+1}}=a_1+\sum\limits_{i=1}^n\left(\frac{i+1}{i}a_i-1\right)= a_1+\sum\limits_{i=1}^{n_0-1}\left(\frac{i+1}{i}a_i-1\right)+\sum\limits_{i=n_0}^{n}\left(\frac{i+1}{i}a_i-1\right) \color{red}{\overset{\text{Pr1}}{\geq}}\\ a_1+\sum\limits_{i=1}^{n_0-1}\left(\frac{i+1}{i}a_i-1\right)+\sum\limits_{i=n_0}^{n}\left(\frac{i+1}{i}-1\right) = a_1+\sum\limits_{i=1}^{n_0-1}\left(\frac{i+1}{i}a_i-1\right)+\color{red}{\sum\limits_{i=n_0}^{n}\frac{1}{i}}$$

and Harmonic series diverges.

---

Proposition 3. If $a_{n_0}\leq\frac{1}{2}$ then $a_n\leq\frac{1}{2},\forall n\geq n_0$.

Results from the induction $$a_{n+1}=\frac{2n+1}{n}a_{n}-1 \leq \frac{2n+1}{n}\frac{1}{2}-1=\frac{1}{2n}\leq \frac{1}{2}$$

---

Now, let's assume the limit exists, then from $(1)$ $$\lim\limits_{n\to\infty}\left(\frac{n+1}{n}a_n-1\right)=0 \iff \lim\limits_{n\to\infty}\frac{n+1}{n}a_n=1 \iff\\ \lim\limits_{n\to\infty}a_n=1$$

and from Pr1, Pr2 and Pr3, $\frac{1}{2}<a_n<1, \forall n$. Now, if

• we want $\frac{1}{2}<a_2<1 $ then $\frac{1}{2}<\frac{3}{1}a_1-1<1 \Rightarrow \color{red}{\frac{1}{2}<a_1<\frac{2}{3}}$
• we want $\frac{1}{2}<a_3<1 $ then $\frac{1}{2}<\frac{5}{2}a_2-1<1 \Rightarrow \frac{6}{10}<a_2<\frac{4}{5}\Rightarrow \frac{6}{10}<\frac{3}{1}a_1-1 <\frac{4}{5} \Rightarrow \color{red}{\frac{16}{30}<a_1< \frac{3}{5}}$
• we want $\frac{1}{2}<a_4<1 $ then $\frac{1}{2}<\frac{7}{3}a_3-1<1 \Rightarrow \frac{9}{14}<a_3<\frac{6}{7} \Rightarrow \\ \frac{9}{14}<\frac{5}{2}a_2-1 <\frac{6}{7} \Rightarrow \frac{46}{70}<a_2< \frac{26}{35} \Rightarrow \\ \frac{46}{70}<\frac{3}{1}a_1-1< \frac{26}{35} \Rightarrow \color{red}{\frac{58}{105}<a_1<\frac{61}{105}}$
• ...
• we want $\frac{1}{2}<a_n<1 $ then $\frac{3(n-1)}{2(2n-1)}<a_{n-1}<\frac{2(n-1)}{2n-1} \Rightarrow \\ \frac{3(n-1)}{2(2n-1)}<\frac{2n-3}{n-2}a_{n-2}-1<\frac{2(n-1)}{2n-1} \Rightarrow \\ \frac{3(n-1)(n-2)}{2(2n-1)(2n-3)}+\frac{n-2}{2n-3}<a_{n-2}<\frac{2(n-1)(n-2)}{(2n-1)(2n-3)}+\frac{n-2}{2n-3} \Rightarrow \\ \frac{3(n-1)(n-2)}{2(2n-1)(2n-3)}+\frac{n-2}{2n-3}<\frac{2n-5}{n-3}a_{n-3}-1<\frac{2(n-1)(n-2)}{(2n-1)(2n-3)}+\frac{n-2}{2n-3} \Rightarrow \\ \frac{3(n-1)(n-2)(n-3)}{2(2n-1)(2n-3)(2n-5)}+\frac{(n-2)(n-3)}{(2n-3)(2n-5)}+\frac{n-3}{2n-5}<a_{n-3}<\frac{2(n-1)(n-2)(n-3)}{(2n-1)(2n-3)(2n-5)}+\frac{(n-2)(n-3)}{(2n-3)(2n-5)}+\frac{n-3}{2n-5} \Rightarrow \\ ... \Rightarrow\\ \frac{3(n-1)!}{2(2n-1)!!} + \frac{(n-2)!}{(2n-3)!!}+...+\frac{1}{3}<a_1<\frac{2(n-1)!}{(2n-1)!!} + \frac{(n-2)!}{(2n-3)!!}+...+\frac{1}{3} \Rightarrow \\ \frac{3(n-1)!}{2(2(n-1)+1)!!} + \frac{(n-2)!}{(2(n-2)+1)!!}+...+\frac{1}{3}<a_1<\frac{2(n-1)!}{(2(n-1)+1)!!} + \frac{(n-2)!}{(2(n-2)+1)!!}+...+\frac{1}{3} \Rightarrow \\ \color{red}{\frac{(n-1)!}{2(2(n-1)+1)!!} +\sum\limits_{k=1}^{n-1}\frac{k!}{(2k+1)!!}<a_1<\frac{(n-1)!}{(2(n-1)+1)!!} +\sum\limits_{k=1}^{n-1}\frac{k!}{(2k+1)!!}}$

What we get is $$\frac{n!}{2(2n+1)!!} +\sum\limits_{k=1}^{n}\frac{k!}{(2k+1)!!}<a_1<\frac{n!}{(2n+1)!!} +\sum\limits_{k=1}^{n}\frac{k!}{(2k+1)!!} \iff\\ \frac{n!}{2(2n+1)!!} -1 +\sum\limits_{k=0}^{n}\frac{k!}{(2k+1)!!}<a_1<\frac{n!}{(2n+1)!!} -1 +\sum\limits_{k=0}^{n}\frac{k!}{(2k+1)!!} \tag{2}$$

We know that (also here) $$\frac{\pi}{2}=\sum\limits_{k=0}^{\infty}\frac{k!}{(2k+1)!!} \text{ and } \lim\limits_{n\to\infty}\frac{n!}{(2n+1)!!}=0$$ thus, by taking the limit from $(2)$ ...

---

Proposition 4. If the limit exists, then $a_1=\frac{\pi}{2}-1$

---

What is left is to check if the sequence converges/diverges for $a_1=\frac{\pi}{2}-1$