Bug in mathematica: computing the sum of the ratios $(n-k+1)/(n-k+2)$ While experimenting with certain sums, I came to the following sum:
$$S_n = \sum_{k=0}^n \frac{n-k+1}{(n+1) (n+2) (n-k+2)}.$$
After rewriting the summand as
$$\frac{n-k+1}{(n+1) (n+2) (n-k+2)} = \frac{n^{\underline{k}}}{(n+2)^{\underline{k}}(n-k+2)^2}$$
and then feeding the sum to Mathematica, i obtained
$$S_n = \frac{7}{6(n+1)(n+2)}$$
But directly summing the first expression above does not yield this closed-form result, because this "result" is wrong! (see edits below) Where is the bug or what am i missing?
EDIT: There seems to be some bug / feature in Mathematica which is leading to the erroneous computation above; the correct answer is of course, as also observed in two answers below. If someone can explain the bug, then he / she should feel free to report it to Wolfram.
(To be specific, the answer above is using $7/6 = (1+2)-H_{1+2}$, instead of $n+2 - H_{n+2}$ as it should)

Here is the same answer with a variation on the calculations 
$\begin{eqnarray*}
S_{n} &=&\sum_{k=0}^{n}\dfrac{n-k+1}{(n+1)(n+2)(n-k+2)} \\
&=&\dfrac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\dfrac{n-k+1}{n-k+2}\qquad (n+1)(n+2)%
\text{ is independent of }k \\
&=&\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\frac{m-1}{m}\qquad \text{substitution }m=n-k+2 \\
&=&\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\left( 1-\dfrac{1}{m}\right)  \\
&=&\dfrac{1}{(n+1)(n+2)}\left(\sum_{m=2}^{n+2}1\right)-\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}%
\dfrac{1}{m} \\
&=&\dfrac{n+2-2+1}{(n+1)(n+2)}-\dfrac{1}{(n+1)(n+2)}\left( \sum_{i=1}^{n+2}%
\dfrac{1}{i}\right) +\dfrac{1}{(n+1)(n+2)} \\
&=&\dfrac{n+1}{(n+1)(n+2)}-\dfrac{H_{n+2}}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)} \\
&=&\dfrac{n+2-H_{n+2}}{(n+1)(n+2)},
\end{eqnarray*}$
where
$H_{n+2}=\displaystyle\sum_{i=1}^{n+2}\dfrac{1}{i}.$
