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)
 A: You can write the sum as
$$\frac1{(n+1)(n+2)} \sum_{k=0}^{n} \frac{n+1-k}{n+2-k},$$
so it's enough to compute the last part of it. This is just
$$\frac{n+1}{n+2} + \frac{n}{n+1} + \dots + \frac{n+1-n}{n+2-n} = 1 - \frac1{n+2} + 1 - \frac1{n+1} + \dots + 1 - \frac1{n+2-n}$$
$$= n + 2 - H_{n+2},$$
where $H_n = \sum_{k=1}^n \frac1{k}$ is the $n$:th Harmonic number.
Your sum is then
$$\frac{n + 2 - H_{n+2}}{(n+1)(n+2)}.$$
A: 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
$H_{n+2}=\displaystyle\sum_{i=1}^{n+2}\dfrac{1}{i}.$
A: Mathematica apparently does not have a nice way of looking at the sum as it is given,
$$S_n = \sum_{k=0}^n \frac{n-k+1}{(n+1) (n+2) (n-k+2)}.$$
However, when you calculate this sum as a definite integral, for both the equation above and the one you derived, ie
$$\sum_{k=0}^{n=p} \frac{n^{\underline{k}}}{(n+2)^{\underline{k}}(n-k+2)^2}.$$
Where p is the number you enter to make it a definite sum, you get the same calculation as when using the original equation.
I initially used the FactorialPower[n,k] command and then I thought that maybe the domain for k was causing issues, so I rewrote the falling factorials using the gamma function to see if that would help. If you are not familiar with how to do this,
$$ n^{\underline{k}} = \frac{ \Gamma (n+1)}{ \Gamma (n-k+1)}$$.
Either way I write the falling factorial, I still come up with the equation you are getting when I use an indefinite sum, which is,
$$S_n = \frac{7}{6(n+1)(n+2)}$$.
As you know, this is in fact not correct. I am not sure why it works for the definite sum, but not the indefinite sum. I will keep thinking about this and let you know what I come up with, if I come up with anything.
If you find an answer, please let us know here. Sorry I cannot be of more help.
