Is there a closed form of this sequence? The sequence is  $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{2}{1}$ , $\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ , $\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ , ...
                   and I need to find n th..
Here's my approach..
I bound them with who has same sum of denominator and numerator.
So $\frac{1}{1}$ is set_1 . 
$\frac{1}{2}$ , $\frac{2}{1}$ are set_2..  
$\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ are set_3 . 
$\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ are set_4 and so on..
If I suppose that n is in set_K , I can conclude n th number is $$ \frac{1+( (n-1)-\sum_{i=1}^K (i-1) ) }{ K - ( (n-1)-\sum_{i=1}^K (i-1) ) } $$
( because set_K starts with $\frac{1}{K}$ and n is ( n - $\sum_{i=1}^K (i-1)$ ) th member of set_K.. ) 
As a result my conclusion is $$\frac{2n-(K-1)K}{K(K+1) - 2(n-1) }$$ but I don't know the relationship between n and K.. 
Is there anyone to help me? I really want to know.. :(

You can break the sequence in groups of $N$ terms with $N=1,2,3,\dots$ of the form
$$\frac{k}{N+1-k}\quad \text {for $k=1,2,\dots ,N$}.$$
So the first $N-1$ groups have $\sum_{i=1}^{N-1}i=\binom{N}{2}$ terms. 
Now, given $n$, there is a unique $N$ such that $\binom{N}{2}<n\leq \binom{N+1}{2}$, which means that the $n$-term is in the $N$-th group and it is equal to
$\frac{k}{N+1-k}$ with $k=n-\binom{N}{2}$. For an explicit formula solve $\binom{N}{2}<n\leq \binom{N+1}{2}$ with respect to $N$:
$$N=\left\lfloor\frac{1+\sqrt{8n-7}}{2}\right\rfloor$$
