<s>[INST] How to simply the sum $\sum\limits_{k=1}^n\frac{N-n+k}{(N-k+1)(N-k+1)(N-k)}$? Let $N>0$ be a large integer, and $n<N$, then how to simply the following sum
$$\sum\limits_{k=1}^n\frac{N-n+k}{(N-k+1)(N-k+1)(N-k)}.$$
Thank you very much, guys.
Actually for another similar sum $\sum\limits_{k=1}^n\frac{1}{(N-k+1)(N-k)}=\sum\limits_{k=1}^n\frac{1}{N-k}-\frac{1}{N-k+1}=\frac{1}{N-n}-\frac{1}{N}$, I know the trick. But adding one term of such thing, $\frac{N-n+k}{N-k+1}$, it becomes difficult. 
So, thanks a million for any clue.
 [/INST] With the same method which you have used above you get 
$$\sum\limits_{k=1}^n \frac{N-n+k}{(N-k+1)^2(N-k)}=\frac{n(2N-n)}{N(N-n)}-(2N+1-n)\sum\limits_{k=1}^n \frac{1}{( N-k+1)^2}$$
Hints:
$\enspace N-n+k=(2N+1-n)-(N-k+1)$
$\enspace \displaystyle \frac{1}{(N-k+1)^2(N-k)}=\frac{1}{N-k}-\frac{1}{N-k+1}-\frac{1}{(N-k+1)^2} $
The closed form for $\sum\limits_{k=1}^n \frac{1}{( N-k+1)^2}$ is:
$$\sum\limits_{k=1}^n \frac{1}{( N-k+1)^2}= \sum\limits_{k=1}^N \frac{1}{k^2} -\sum\limits_{k=1}^{N-n} \frac{1}{k^2}=$$$$((\frac{1}{N!} \begin{bmatrix} N+1 \\ 2 \end{bmatrix})^2-\frac{2}{N!} \begin{bmatrix} N+1 \\3 \end{bmatrix})-((\frac{1}{(N-n)!} \begin{bmatrix} N-n+1 \\ 2 \end{bmatrix})^2-\frac{2}{(N-n)!} \begin{bmatrix} N-n+1 \\3 \end{bmatrix}) $$
where $\begin{bmatrix} n \\ k \end{bmatrix}$ is called unsigned Stirling number of the first kind . 
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