Can this expression be simplified? I get a function related to a variable $x$, that is 
$f(x)=\displaystyle \frac{ \sum_{i=0}^K{ i \binom{n+i}{i} x^i}}
{\sum_{i=0}^K{\binom{n+i}{i}x^i}}$,
where $K$ and $n$ are positive integers, can this expression be simplified to some other expression?
 A: Here is my best shot. Define $G$ and $F$ as 
$$ f(x) = \frac{G(x)}{F(x)} $$
Now let's do away with the $K$ and work in the ring of formal power series, defining the analogs $\hat{f}, \hat{G}, \hat{F}$. We notice the denominator can be expressed as the generating function
$$ \hat{F}(x) \;\; = \;\; \sum_{k\geq0} {n+k \choose k} x^k \;\; = \;\; \sum_{k\geq0} {-n-1 \choose k} (-x)^k \;\; = \;\; (1-x)^{-n-1} $$  
The numerator is also formally just the shifted term-wise "derivative" of the denominator
$$  \hat{G}(x) \;\; = \;\; \sum_{k\geq0} k {n+k \choose k} x^k \;\; = \;\; x \, \hat{F}\,'(x) \;\; = \;\; (n+1) \, x \, (1-x)^{-n-2} $$  
Dividing the two (which is legal since $\hat{F}(0)=1$) yields
$$ \hat{f}(x) \;\; = \;\; \frac{(n+1) \, x}{1-x} \;\; = \;\; (n+1) \, (x + x^2 + x^3 \; + \;\; ... ) $$
Now what does this mean? This means that in the limit $K \rightarrow \infty$ we have $f(x) \rightarrow \hat{f}(x)$. But how well do $f$ and $\hat{f}$ agree for finite $K$? Let's take a look at $\hat{f}$ again  
$$ \begin{align}
\hat{f} \;\; = \;\; \frac{\hat{G}}{1-(1-\hat{F})} \;\; &= \;\; \hat{G} \left[ 1 + (1-\hat{F}) + (1-\hat{F})^2 \; + \;\; ... \right]  \\
&= \left[ G + \mathcal{O}\left(x^{K+1}\right) \right] \left [ 1 + (1-F) + (1-F)^2 \; + \;\; ... \; + \; \mathcal{O}\left(x^{K+1}\right) \right] \\ & \\
&= f + \mathcal{O}\left(x^{K+1}\right)
\end{align}$$  
So the two series agree up thru the coefficient of $x^K$. Our final result is  
$$ \begin{align}
f(x) &= \sum_{k=1}^K (n+1) \, x^k &+ \mathcal{O}\left(x^{K+1}\right) \\
&= (n+1) \,\frac{x\,(x^K-1)}{x-1} &+ \mathcal{O}\left(x^{K+1}\right) 
\end{align}$$
