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.
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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\begin{align}
&\color{#f00}{%
\sum_{k = 1}^{n}{N - n + k \over \pars{N - k + 1}\pars{N - k + 1}\pars{N - k}}}
\\[5mm] = &\
\pars{2N - n}\pars{\sum_{k = 1}^{n}{1 \over k - N - 1} -
\sum_{k = 1}^{n}{1 \over k - N}} -
\pars{2N - n + 1}\sum_{k = 1}^{n}{1 \over \pars{k - N - 1}^{2}}
\\[5mm] = &\
{\pars{2N - n}n \over N\pars{N - n}} -
\pars{2N - n + 1}\sum_{k = 1}^{n}{1 \over \pars{k - N - 1}^{2}}
\end{align}

Note that

$$\left\{\begin{array}{rcl}
\ds{\sum_{k = 1}^{n}{1 \over k + a}} & \ds{=} &
\ds{\sum_{k = 0}^{n - 1}{1 \over k + 1 + a} =
\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 + a} - {1 \over k + n + 1 + a}} =
H_{n + a} - H_{a}}
\\[2mm]
\ds{\sum_{k = 1}^{n}{1 \over \pars{k + a}^{2}}} & \ds{=} &
\ds{\partiald{}{a}\pars{H_{a} - H_{n + a}} = \Psi\,'\pars{1 + a} -
\Psi\,'\pars{1 + n + a}}
\end{array}\right.
$$

$\ds{H_{z}}$ is the Harmonic Number and $\ds{\Psi\,'}$ is the Trigamma Function.


Then,
\begin{align}
&\color{#f00}{%
\sum_{k = 1}^{n}{N - n + k \over \pars{N - k + 1}\pars{N - k + 1}\pars{N - k}}}
\\[5mm] = &\
\color{#f00}{{\pars{2N - n}n \over N\pars{N - n}} -
\pars{2N - n + 1}\bracks{\Psi\,'\pars{-N} - \Psi\,'\pars{-N - n}}}
\end{align}

Any issue with the Trigamma's argument signs can be deal with the Euler Reflection Formula or/and the Recurrence Formula:

$$
\left\{\begin{array}{rcl}
\ds{\Psi\,'\pars{z}} & \ds{=} &
\ds{-\Psi\,'\pars{1 - z} + \pi^{2}\csc^{2}\pars{\pi z}}
\\[2mm]
\ds{\Psi\,'\pars{z + 1}} & \ds{=} & \ds{\Psi\,'\pars{z} - {1 \over z^{2}}}
\end{array}\right.
$$
A: 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 . 
