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$$\sum_{n = 1}^{D - 1} \frac{n}{D - n}$$

Is it possible to reduce this summation to a function of $D$?

amWhy
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bestscammer5
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    Let $$f(D) = \sum_{n = 1}^{D - 1} \frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation? – Alex Vong Nov 28 '18 at 12:22

1 Answers1

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$$\sum_{n = 1}^{D - 1} \frac{n}{D - n}\\ =\sum_{n=1}^{D-1}\left( -1 + \frac{D}{D-n}\right)\\ =-D+1 + D\sum_{n=1}^{D-1}\frac1{D-n} \\ =-D+1 + D\sum_{m=1}^{D-1}\frac1{m} \\ = -D+1 + D H_{D-1}$$ where $H_n$ is the $n$th harmonic number.

Calvin Khor
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