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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)

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    You sure about what you're getting? I get $\frac1{n+1}-\frac1{n^2+3n+2}\sum_{j=1}^{n+2}\frac1{j}$.2010-11-21
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    @JM: I tried summing with: FactorialPower[n,k]/(FactorialPower[n+2,k] (n-k+2)^2 and that gives me the simplied sum i wrote; otherwise after replacing $n-k$ by $k$ in the first sum, I got something like what you have. i am confused as of now!2010-11-21
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    `$VersionNumber`, if you don't mind?2010-11-21
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    am using version 7.2010-11-21
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    `FactorialPower[n,k]/(FactorialPower[n+2,k] (n-k+2)^2` seems to be missing needed parentheses...2010-11-21
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    @JM: sorry for the typo; luckily, in mma, 2-D input makes such typos impossible ;-)2010-11-21
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    For pasting into places like these, `InputForm[]` helps a great deal. ;) In any event, there really is [something screwy](http://tinyurl.com/2fbzzk9) going on...2010-11-21
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    Yes, indeed, which is what threw me off. thanks for taking the time to look at this thing. i will look again more carefully, and also check back here in case you have found the bug / misunderstanding that i am having!2010-11-21
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    On a lark, try re-expressing your falling factorials as Pochhammer symbols, and have *Mathematica* at it again. I'm getting an `Indeterminate` in 5.2, but maybe the behavior is now different in your version.2010-11-21
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    I think, either this is a bug or a "feature", but i am not expert enough with mathematica to judge whether this is a bug or a feature though. $7/6$ is 3-HarmonicNumber[3], so somehow mathematica is replacing $n$ by $1$, and this might be a bug.2010-11-22
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    You should ask on MathGroup, which you can access for example through googlegroups. That's the correct place for this question.2010-11-23
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    A debugging suggestion: try using variable upper and lower limits; thus `Sum[(* something *), {k, l, n}]` and then `% /. l -> 0` (or probably a `Limit[]` if need be.)2010-11-24
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    i am voting to close this question; but I dont want to delete it. could others please vote to close it?2010-11-24

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