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)