Given a sequence $a_{n}= \frac{2\,a_{n+ 1}}{1+ a_{n+ 1}^{2}}\in (0, 1).$ Prove that $\sum a_{n}< \frac{2\,a_{1}}{1- a_{1}}.$

Question

Given a sequence $a_{n}= \dfrac{2\,a_{n+ 1}}{1+ a_{n+ 1}^{2}}\in (0, 1).$ Prove that $\sum a_{n}< \dfrac{2\,a_{1}}{1- a_{1}}.$

How can I seperate $n$ parts from $\dfrac{2\,a_{1}}{1- a_{1}}$ and each of them $u_{n}> a_{n}{\it ?}$ I observe $$a_{n}- a_{n+ 1}= \dfrac{a_{n+ 1}\left ( 1- a_{n+ 1}^{2} \right )}{1+ a_{n+ 1}^{2}}= \dfrac{a_{n}\left ( 1- a_{n+ 1}^{2} \right )}{2}$$ So, what $u_{n}- u_{n+ 1}$ sould be ? I need to the help, thanks a real lot.

Edit: I found one relationship that is $$\frac{2\,a_{n+ 1}}{1+ a_{n+ 1}^{2}}\geq a_{n+ 1}\left ( 2- a_{n+ 1} \right )$$ Sorry I can't talk it clearly now, because it's too late here. I'll post the proving of the inequality as soon as well.

Answer 1

Using the double-angle formula for the hyperbolic tangent one can obtain an explicit formula for $a_n$ and get the desired inequality:

We know that $a_n \in (0, 1)$, so that we can substitute $a_n = \tanh(x_n)$ with $x_n > 0$. Then $$ \tanh(x_n) = \frac{2 \tanh(x_{n+1})}{1+\tanh^2(x_{n+1})} = \tanh(2 x_{n+1}) $$ implies $x_n = 2x_{n+1}$. It follows that $x_n = x_1/2^{n-1}$ for all $n$. Now the sum can be estimated: $$ \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \tanh\left(\frac{x_1}{2^{n-1}}\right) < \sum_{n=1}^\infty \frac{x_1}{2^{n-1}} = 2 x_1 \, . $$ On the other hand, $$ \frac{2a_1}{1-a_1} = \frac{2\tanh(x_1)}{1-\tanh(x_1)} = e^{2x_1} - 1 > 2x_1 \, . $$

Answer 2

First, you need to prove that $a_n$ converges to $0$. With what you wrote, you can notice that $\forall n, a_{n+1}\leq a_n $ and $a_n>0 $.

Hence, $a_n$ is monotone and bounded, so it converges. Since you have $f(a_n)=a_{n+1} $ with $f$ the bijection of $\frac{2x}{1+x^2} $, it must converge to a fixed point of $f$, and you can notice that $0$ is one.

(edit :) $1$ is also a fixed point of $f$, however, since $a_1<1$ and $a_n$ is decreasing, it can't converge to $1$ : $$\lim_{n\to\infty} a_n\leq a_1 <1 $$

So you get that $a_n$ converges to $0$, but it's not over yet !

You can use telescopic sums to get :

$$\sum_{n=1}^{N-1} (a_{n+1}-a_n )=a_N-a_1=\sum_{n=1}^{N-1}\frac{a_n(a_{n+1}^2-1)}{2}$$ since you have $a_n$ decreasing and less than $1$, you know that : $$a_{n+1}^2\leq a_{n+1}\leq a_n\leq ...\leq a_1 $$

Hence,$ \frac{a_n(a_{n+1}^2-1)}{2}\leq \frac{a_n(a_1-1)}{2} $

So,

$$a_N-a_1\leq\frac{(a_1-1)}{2} \sum_{n=1}^{N-1}a_n\Leftrightarrow\\ \frac{2a_N}{a_1-1}-\frac{2a_1}{a_1-1}\geq \sum_{n=1}^{N-1}a_n \text{ (because $a_1<1$)}$$

An now you can take the limit, since $a_n$ goes to $0$, and $\sum a_n $ is an increasing sequence and it is bounded, so it converges, and you get :

$$\boxed{\sum a_n\leq \frac{2a_1}{1-a_1} }$$