What method can I use to compute the limit of this series? Let
$$
\begin{cases}
a_1 &=1 \\a_{n+1}&=\frac{1}{a_1+a_2+\cdots +a_n}-\sqrt{2}
\end{cases}
$$
Then $\sum_{i=1}^{\infty}{a_i}=?
$
I assume that the limit exists, and then I get the limit is equal to $\frac{\sqrt{2}}{2}$
$\sum_{k=1}^{n}{a_k=A_n,}A_{n+1}-A_n=\frac{1}{A_n}-\sqrt{2}.$take the limit of both sides can get $\frac{\sqrt{2}}{2}$.
$$A_n+\frac{1}{A_n}-\frac{3}{\sqrt{2}}=\frac{\left( \sqrt{2}A_n-1 \right) \left( A_n-\sqrt{2} \right)}{\sqrt{2}A_n}=\frac{\sqrt{2}A_{n+1}-1}{\sqrt{2}}$$
$$\frac{\left( \sqrt{2}A_n-1 \right) \left( A_n-\sqrt{2} \right)}{A_n}=\sqrt{2}A_{n+1}-1=\left( \sqrt{2}A_n-1 \right) \left( 1-\frac{\sqrt{2}}{A_n} \right)$$
All I have to do is prove that $\left( \frac{\sqrt{2}}{A_n} \right)$  has a bound greater than 0 and less than 1, but I didn't prove that bound.
 A: We recall that   $\ A_n= \displaystyle \sum_{i=1}^n a_i$.
We have: $\ A_1 = 1 \ \text{ and } \ \forall n \in \mathbb N \ , \ A_{n+1}=A_n+\dfrac{1}{A_n}-\sqrt{2}$
Let  $\ f(x)=x+\dfrac{1}{x}-\sqrt{2}$.
We have:
$f\circ f(x)-\dfrac{\sqrt{2}}{2} = \left(x-\dfrac{\sqrt{2}}{2}\right) \dfrac{ (\sqrt{2}-x)(1+\sqrt{2}-x)(x+1-\sqrt{2})}{x(x^2-\sqrt{2}x+1)}$
$f\circ f(x)-x = - \left( x-\dfrac{\sqrt{2}}{2}\right)^3 \dfrac{2\sqrt{2}}{x^2-\sqrt{2}x+1}$
So:  $\ \forall x \in \left[ \dfrac{\sqrt{2}}{2},1\right] \ , \ \dfrac{\sqrt{2}}{2} \leqslant f(x)\leqslant x \leqslant 1$
Then, by induction: $\ \forall n \in \mathbb N \ , \ \dfrac{\sqrt{2}}{2} \leqslant A_{2n+1}\leqslant 1$
And: $\forall n \in \mathbb N \ , \ A_{2n+3} \leqslant A_{2n+1}$
$(A_{2n+1})_{n\in \mathbb N}$ is decreasing and bounded. It is convergent vers a real $\ell$ and $\ell = f\circ f)(\ell)$. So $\ \ell=\dfrac{\sqrt{2}}{2}$.
Now $(A_{2n+2})_{n\in\mathbb N} =(f(A_{2n+1})_{n\in \mathbb N})$ converges towards $f(\ell)=\ell$.
We conclude that $(A_n)_{n\in\mathbb N^*}$ converges towards $\ell=\dfrac{\sqrt{2}}{2}$
