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I am wondering if there is a closed form expression for the sum \begin{align} \sum_{n=1}^{N-1} \frac{n}{N-n}. \end{align}

We note that there is a symmetry \begin{align} \sum_{n=1}^{N-1} \frac{n}{N-n} = \sum_{n=1}^{N-1}\frac{N-n}{n} \end{align} which might be helpful. One observation which I do not know may be helpful is that this can be seen as a special case of a general sum with Mobius conformal mappings \begin{align} z\rightarrow \gamma z=\frac{az+b}{cz+d}, \quad \gamma \in SL(2,\mathbb{C}), \quad \sum_{z \in \{1,\cdots,N-1\}} \gamma z \end{align} where $\gamma=\begin{pmatrix}N & -1 \\ 1 & 0\end{pmatrix}$. However, since the Mobius map is not linear, this may not be good way to start.

Please let me know if anyone knows the answer.

Nugi
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    I assume you mean the sum to be over $n$ and not $i$ (which doesn't appear in the summand)? – jwimberley Jan 21 '21 at 15:48
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    The result is clearly equal to $N H_{N-1} - N + 1$, which is of course not closed form, but this reduces your question to asking if there is a closed form expression for the harmonic numbers $H_N$. – jwimberley Jan 21 '21 at 15:51
  • A good approximation for $H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ is $H_{n}\approx \ln(n) + \gamma$, where $\gamma\approx 0.577$. This approximation gets better as $n$ increases. – Joshua Wang Jan 21 '21 at 15:56
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    I believe (https://math.stackexchange.com/questions/52572/do-harmonic-numbers-have-a-closed-form-expression) and links thereof shows that it is proven that $H_N$ has no expression which is rational function in $N$. – Nugi Jan 21 '21 at 16:02
  • @Nugi: this also follows immediately from the asymptotics (in the preceding comment). – metamorphy Jan 21 '21 at 16:05
  • @metamorphy Perhaps how good the approximation is relevant for that approximation formula to imply that there is no expression as a rational function in N? – Nugi Jan 21 '21 at 16:16
  • @Nugi: more precisely, there's no rational $f(N)$ with $\lim\limits_{N\to\infty}f(N)/\log N=1$. – metamorphy Jan 21 '21 at 16:42

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