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The $n$-th polytopic number of $r$ dimensional simplex is given by

$$P_r(n)=\begin{pmatrix}n+r-1 \\r \end{pmatrix}= \frac{(n+r-1)!}{r!(n-1)!} =\frac{n^{\bar{r}}}{r!}$$

Based on that, how do we proof that

$$\sum_{i=1}^n {\frac{1}{P_r(i)}}={\frac{r}{r-1}}\Big(1-{\frac{1}{P_n(r)}}\Big)$$

Do correct me if I'm wrong, I based it from paper by O. Ufuoma (http://eprints.southasianlibrary.com/id/eprint/507/1/Ufuoma1422019ARJOM48075.pdf).

Unfortunately, the proof is not included on that manuscript.

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For $n = 1$, $$\frac{1}{P_r(1)} = \begin{pmatrix} r \\ r \end{pmatrix}^{-1} = 1 = \frac{r}{r-1}(1 - r^{-1}) = \frac{r}{r-1}\left(1 - \begin{pmatrix} r \\ 1 \end{pmatrix}^{-1}\right) = \frac{r}{r-1}\left(1 - \frac{1}{P_1 (r)} \right)$$

Now suppose the statement is true for $n = m$. Then for $n = m+1$, we get \begin{align} \sum_{i = 1}^{m+1} \frac{1}{P_r(i)} &= \sum_{i = 1}^{m} \frac{1}{P_r(i)} + \frac{1}{P_r(m+1)}\\ &= \frac{r}{r-1}\left(1 - \frac{1}{P_m(r)}\right) + \frac{r! m!}{(r + m)!}\\ &= \frac{r}{r-1}\left(1 - \frac{m!(r-1)!}{(r+m -1)!} + \frac{(r-1)(r-1)!m!}{(r+m)!} \right)\\ &= \frac{r}{r-1}\left(1 - \frac{m! (r-1)! (r+m) - (r-1)(r-1)!m!}{(r+m)!}\right)\\ &= \frac{r}{r-1}\left(1 - \frac{m! (r-1)! (r+m - r +1)}{(r+m)!}\right)\\ &= \frac{r}{r-1}\left(1 - \frac{(m+1)! (r-1)!}{(r+m)!}\right) = \frac{r}{r-1}\left(1 - \frac{1}{P_{m+1}(r)}\right) \end{align}

Hence via induction, we are done.

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