Value of $\sum\limits_{n=1}^\infty a_{n}$ if $a_1=1$ and $a_{n+1}=\frac{1}{a_{1}+a_{2}+....+a_{n}}-\sqrt2$ for every $n\ge1$ 
The sequence $(a_n)$ is defined by setting $a_{1}=1$ and, for every $n\ge1$, $$a_{n+1}=\frac{1}{a_{1}+a_{2}+....+a_{n}}-\sqrt2.$$
  Find the sum of the series $\sum\limits_{n=1}^\infty  a_{n}$.

A hint on how to begin solving this will do. Should I try factoring it out somehow?
 A: Let we try to settle down what is stated in the comments above. If we set:
$$ A_n = a_1+a_2+\ldots+a_n \tag{1}$$
the recursion can be written as:
$$ A_{n+1}-A_n=\frac{1}{A_n}-\sqrt{2},\tag{2}$$
hence if we assume that $\lim_{n\to +\infty}A_n = L$, we must have $L=\frac{1}{\sqrt{2}}$.
So it is enough to prove that $\lim_{n\to +\infty}A_n$ exists. If we set $B_n=A_n\sqrt{2}$, $(2)$ gives:
$$ B_{n+1}-B_n=2\left(\frac{1}{B_n}-1\right) \tag{3}$$
or:
$$ B_{n+1}B_n = 1+(B_n-1)^2\tag{4} $$
so it is enough to study the iteration $x\mapsto \frac{1+(x-1)^2}{x}$ with starting point $x_0=\sqrt{2}$.
The function $f(x)=\frac{1+(x-1)^2}{x}$ over the interval $I=\left[2\sqrt{2}-2,\sqrt{2}\right]$ is convex and decreasing. $x=1$ is the only solution of $f(x)=x$ over $I$, and $|f'(x)|$ over $I$ is bounded by $\frac{1}{2}+\sqrt{2}$.
We may also check that $f(f(x))$ is a contraction mapping on $I$, since over $I$ we have $\left|\frac{d}{dx}f(f(x))\right|\leq 1$ and equality is attained only at $x=1$, hence the Banach fixed-point theorem gives that $B_n\to 1$, i.e. $A_n\to\frac{1}{\sqrt{2}}$, as wanted.
