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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

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Assuming the sum $S = \sum_{n=1}^\infty$ exist, by your equation it must be that $$S = \lim_{n \to \infty} S_{n+1} = \lim_{n \to \infty} (S_n + 1/S_n - \sqrt{2}) = S + 1/S - \sqrt{2}.$$ So $S = 1/\sqrt{2} = \frac{1}{2}\sqrt{2}$.This of course leaves the proof that the limit really exists. –  Magdiragdag Sep 6 '13 at 19:03

1 Answer 1

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

$$ 1=\limsup\frac{|b_{n+1}|}{|b_n|}=\limsup|1-\frac{\sqrt{2}}{b_n+\frac{1}{\sqrt{2}}}|<1 $$

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Thank you, it's very nice. –  math110 Sep 7 '13 at 9:40

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