Solution to this recurrence? Is there exists a solution to this recurrence.
$$F(N,1) = N, N≥1$$
$$F(N,K) = \frac {1}{\lfloor\frac 1{F(N-1,K-1)} -\frac 1{F(N,K-1)}\rfloor} \;\;\;\;2≤K≤N$$
I tried to simplify the equation but i am not able to find F(1,2) and thus unable to proceed.
I am new to this site,so please let me know if i have not asked the question properly. 

The original recurrence relation is equivalent to
$$
\frac{1}{F(N,K)}
=
\frac{1}{F(N-1,K-1)}
-
\frac{1}{F(N,K-1)}.
$$
For $K = 2$, we have
\begin{align}
\frac{1}{F(N,2)}
&=
\frac{1}{F(N-1,1)}
-
\frac{1}{F(N,1)} \\
&=
\frac{1}{N-1}
-
\frac{1}{N}
\\
&=
\frac{1}{N\,(N-1)}.
\end{align}
Next, for $K = 3$, we have
\begin{align}
\frac{1}{F(N,3)}
&=
\frac{1}{F(N-1,2)}
-
\frac{1}{F(N,2)} \\
&=
\frac{1}{(N-1) \, (N-2)}
-
\frac{1}{N \, (N-1)}
\\
&=
\frac{N}{N \, (N-1) \, (N-2)}
-
\frac{N-2}{N \, (N-1) \, (N-2)}
\\
&=
\frac{2}{N\,(N-1)\,(N-2)}.
\end{align}
So we can guess,
(1)
\begin{align}
\frac{1}{F(N,K)}
&=
\frac{(K-1)!}{N\,(N-1)\,(N-2)\,(N-K+1)}.
\end{align}
Let us prove it by induction,
suppose (1) holds for $K = k$,
For $K = k+1$,
\begin{align}
\frac{1}{F(N,k+1)}
&=
\frac{1}{F(N-1,k)}
-
\frac{1}{F(N,k)} \\
&=
\frac{(k-1)!}{(N-1) \, (N-2) \, (N-k)}
-
\frac{(k-1)!}{N \, (N-1) \, (N-k+1)}
\\
&=
\frac{(k-1)! \, N }{ N \, (N-1) \, (N-2) \dots (N-k)}
-
\frac{(k-1)! \, (N-k)}{N \, (N-1) \, (N-k+1)\dots (N-k)}
\\
&=
\frac{k!}{N\,(N-1)\,(N-2)\,(N-k)},
\end{align}
which means (1) also holds for $K = k+1$.
So
$$
F(N,K) = \frac{N(N-1)\dots(N-K+1)}{(K-1)!} = N \, {N-1 \choose K-1}.
$$
