Limit in the set of real sequences. I have troubles trying to prove the following proposition:
Let $S$ be the set of real sequences with $d(\tilde{x},\tilde{y})=\displaystyle\sum_{i=1}^{\infty}\frac{|x_{i}-y_{i}|}{2^{i}(1+|x_{i}-y_{i}|)}.$ 
Then $d(\tilde{x}^{k},\tilde{x})\rightarrow 0$ when $k\rightarrow\infty$ $\Leftrightarrow$ $\tilde{x}_{i}^{k}\rightarrow x_{i}$ when $k\rightarrow\infty$ For all $i\in\mathbb{N}$
where $\tilde{x}^{k}=(\tilde{x}_{i}^{k}),\tilde{x}=(x_{i})\in S.$ 
My attempt was to use the definition of convergence of sequences, but only achive that $\frac{|x_{i}^{k}-x_{i}|}{2^{i}(1+|x_{i}^{k}-x_{i}|)}<\epsilon.$ 
But I cannot see how it helps me to prove $|x_{i}^{k}-x_{i}|<\epsilon.$
The other implication  think already I get it.
Any kind of help would be greatly appreciated.
 A: Suppose not. Then exists a $\epsilon>0
 $ and exists a $\overline{i}\in\mathbb{N}
 $ such that $$ \left|x_{\overline{i}}^{k}-x_{\overline{i}}\right|\longrightarrow\epsilon
 $$ as $k\rightarrow\infty
 $ then $$\sum_{i\geq0}\frac{\left|x_{i}^{k}-x_{i}\right|}{2^{i}\left(1+\left|x_{i}^{k}-x_{i}\right|\right)}=\frac{\left|x_{\overline{i}}^{k}-x_{\overline{i}}\right|}{2^{\overline{i}}\left(1+\left|x_{\overline{i}}^{k}-x_{\overline{i}}\right|\right)}+\sum_{i\geq0,i\neq\overline{i}}\frac{\left|x_{i}^{k}-x_{i}\right|}{2^{i}\left(1+\left|x_{i}^{k}-x_{i}\right|\right)}
 $$ and so if we take the limit as $k\rightarrow\infty
 $ we have $$0=\frac{\epsilon}{2^{\overline{i}}\left(1+\epsilon\right)}+C
 $$ where $C\geq0
 $ and this is absurd.
A: Hint.
Look at the behavior of the function $$f(x)=\frac{x}{1+x}$$ for $x \ge 0$. In particular, where is it close to zero?
A: let ${ x }_{ n }=\left\{ { \xi  }_{ i }^{ n } \right\}$  ,$x=\left\{ { \xi  }_{ i\quad  } \right\}$  and  $x_{ n }\rightarrow x\quad it\quad$ means  that
 $\sum _{ i=1 }^{ \infty  }{ \frac { \left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  }{ 2^{ i }\left( 1+\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  \right)  }  } <\varepsilon$  in case $ n\ge { n }_{ 0 }(\varepsilon )$
in every fixed i we have got $\frac { \left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  }{ 2^{ i }\left( 1+\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  \right)  } <\varepsilon$ in case $ n\ge { n }_{ 0 }(\varepsilon ) $ and i is fixed and $\varepsilon $ is arbitrary then,
$\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right| \rightarrow 0,\quad$ when $\quad n\rightarrow \infty$
lets inverse, 
$$\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right| \rightarrow 0,\quad  when \quad n\rightarrow \infty$$
for every i.Take arbitrary number $\varepsilon $.First we choose m for
$\sum _{ i=m+1 }^{ \infty  }{ \frac { 1 }{ 2^{ i } }  } <\frac { \varepsilon  }{ 2 } $
  then 
$$
d({ x }_{ n },x)=\sum _{ i=1 }^{ \infty  }{ \frac { 1 }{ 2^{ i } } \frac { \left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  }{ 1+\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  }  } =\sum _{ i=1 }^{ m }{ \frac { 1 }{ 2^{ i } } \frac { \left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  }{ 1+\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  } + } \sum _{ i=m+1 }^{ \infty  }{ \frac { 1 }{ 2^{ i } } \frac { \left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  }{ 1+\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  } < } \sum _{ i=1 }^{ m }{ \frac { 1 }{ 2^{ i } } \frac { \left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  }{ 1+\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  } + } \frac { \varepsilon  }{ 2 } $$ 
due to the fact that ,the last sequence is finit ,so $\sum _{ i=1 }^{ m }{ \frac { 1 }{ 2^{ i } } \frac { \left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  }{ 1+\left| { \xi  }_{ i }^{ n }-\xi _{ i } \right|  } < } \frac { \varepsilon  }{ 2 } $ where $ n\ge { n }_{ 0 }(\varepsilon )$?we can write $d({ x }_{ n },x)<\varepsilon $
but if you want to show that S is a metric space with this above metrica,first two axioms are obvious,third one you can prove with this inequetion 
$ \frac { \left| a+b \right|  }{ 1+\left| a+b \right|  } \le \frac { \left| a \right|  }{ 1+\left| a \right|  } +\frac { \left| b \right|  }{ 1+\left| b \right|  } $
