How find this $\sum_{n=1}^{\infty}a_{n}$ let $$a_{1}=1,a_{n+1}=\dfrac{1}{a_{1}+a_{2}+\cdots+a_{n}}-\sqrt{2}$$
find the value
$$\sum_{n=1}^{\infty}a_{n}$$
my try:let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$,then we have
$$S_{n+1}=S_{n}+a_{n+1}=S_{n}+\dfrac{1}{S_{n}}-\sqrt{2}$$
and I find $S_{1}=1,S_{2}=2-\sqrt{2},S_{3}=3-\dfrac{3}{2}\sqrt{2}$
I guess
$$S_{1}>S_{3}>\cdots>S_{2n-1},S_{2}>S_{4}>\cdots>S_{2n}$$
so I try prove
$$S_{2n+1}-S_{2n-1}<0$$
if $n$ is odd. then
$$S_{n+2}=S_{n+1}+\dfrac{1}{S_{n+1}}=S_{n}+\dfrac{1}{S_{n}}-\sqrt{2}+\dfrac{1}{S_{n}+\dfrac{1}{S_{n}}-\sqrt{2}}-\sqrt{2}$$
let $$f(x)=x+\dfrac{1}{x}-\sqrt{2}\Longleftrightarrow S_{n+2}=f(S_{n+1})=f(f(S_{n}))$$
where $x=S_{n}$
and $$S_{n+2}-S_{n}=\dfrac{1}{x}-\sqrt{2}+\dfrac{1}{x+\dfrac{1}{x}-\sqrt{2}}-\sqrt{2}=\dfrac{(1-\sqrt{2}x)^3}{x(x^2-\sqrt{2}x+1)}$$
so if $x\in \left(\dfrac{1}{\sqrt{2}},1\right)$, then we have $$S_{n+2}<S_{n}$$
where $n$ is odd numbers.
my question: How can I  determine
 $x=S_{n}$(n is odd) in $(\dfrac{1}{\sqrt{2}},1)$?
and I think this problem have other nice methods.Thank you 

Very nice problem. Here is my approach: 
If the series converges, it converges to $\frac{1}{\sqrt{2}}$ since  $a_n\rightarrow 0$. Taken in consideration this, we define $$b_n=a_1+\cdots+a_n-\frac{1}{\sqrt{2}}$$
We want to prove that $b_n$ converges (to zero).
We have
$$
b_{n+1}=b_n+a_{n+1}=b_n-\sqrt{2}+\frac{1}{b_n+\frac{1}{\sqrt{2}}}=b_n-\frac{\sqrt{2}b_n}{b_n+\frac{1}{\sqrt{2}}}
$$
Note that if $b>0$ then
$$
0<\frac{ \sqrt{2}b}{b+\frac{1}{\sqrt{2}}}<2b
$$
Note also that
$$
b-\frac{ \sqrt{2}b}{b+\frac{1}{\sqrt{2}}}>\frac{-1}{\sqrt{2}}
$$
For the other side, if $-\frac{1}{\sqrt{2}}<b<0$ then
$$
0>\frac{ \sqrt{2}b}{b+\frac{1}{\sqrt{2}}}>2b
$$
It implies that if $b_n>0$ then $b_n>b_{n+1}>-b_n$. In same way, if $-\frac{1}{\sqrt{2}}<b_n<0$, then $b_n<b_{n+1}<-b_n$. These inequalities can be written in compact form as:
$$
|b_{n+1}|<|b_n|
$$
and therefore $\{|b_n|\}$ converges to $L$. It only remains to prove that $L=0$.
For this, note that if $\frac{1}{\sqrt{2}}>L>0$, then
$$
1=\limsup\frac{|b_{n+1}|}{|b_n|}=\limsup|1-\frac{\sqrt{2}}{b_n+\frac{1}{\sqrt{2}}}|<1
$$
