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Fix a positive integer $m$. I am trying to prove \[\sum \frac 1 {k_1\cdots k_m} = \sum_{k=1}^n \frac{(-1)^{k-1}}{k^m} \binom n k\] where the sum on the left is taken over all integer $m$-tuples $(k_1,\dots,k_m)$ such that $1\le k_1\le \cdots\le k_m\le n$. This comes immediately after having proved that \[ \sum_{k=1}^n b_k/k = \sum_{k=1}^n \frac{a_k}k \binom n k \] where $b_n = \sum_{k=1}^n a_k\binom n k$. There's a strong similarity, so I suspect that I should use this result to prove the above. In particular it suggests that we take $a_k/k = (-1)^{k-1}/k^m$ which is to say $a_k = (-1)^{k-1}/k^{m-1}$. If this is the right path then we want to show that \[ \sum\frac 1 {k_1\cdots k_m} = \sum_{k=1}^n \frac{b_k}k = \sum_{k=1}^n \frac 1 k \sum_{j=1}^k a_j\binom{k}j = \sum_{k=1}^n \frac 1 k \sum_{j=1}^k \frac{(-1)^{j-1}}{j^{m-1}}\binom{k}j\] Now trying to rearrange and reindex the sums basically just seems to get you back into the proof that $\sum b_k/k = \sum a_k\binom n k /k$ so that seems like a dead end.

I'm guessing that somehow, instead, we should find a more direct reason why these are equal, perhaps by splitting fractions. For instance, $\frac 1 {2\cdot 3} = \frac{1}{2}-\frac{1}{3}$ but there is no such simple splitting for $\frac 1 {2\cdot 4}$.

The right-hand side of the above seems to suggest that perhaps we fix a $1\le k \le n$ as the least number in $(k_1,\dots,k_m)$, and then use something that looks vaguely like inclusion-exclusion.

From here the question looks like "Is there a reason why, if we collect all the terms of $\sum \frac 1 {k_1\cdots k_m}$ with a fixed $k_1$, that the sum of these terms is $\frac 1 k \sum_{j=1}^k \frac{(-1)^{j-1}}{j^m}\binom k j$?"

Just to test the idea out, if we set $n=4,m=2$ and $k_1=2$ then we would be looking at the terms \[ \frac{1}{2\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 4} \] and on the other hand we would be trying to see whether this is the same as \[ \frac 1 2 \sum_{j=1}^2 \frac{(-1)^{j-1}}{j^2}\binom 2 j = \frac 1 2 \cdot \left( 2 - \frac 1 4 \right)\] A quick computation shows these are not equal, so then I'm not sure if I made a mistake along the way, or if the entire plan of attack is flawed.

If the entire plan is flawed, I don't have a plan B.

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  • $\begingroup$ In that third equation the final expression on the RHS inside the first summation is the same as the RHS of the identity you want to prove but with $m$ replaced by $m-1$. That suggests an induction on $m$; have you tried this? $\endgroup$ Commented Sep 17, 2022 at 18:59
  • $\begingroup$ This is a very nice looking problem. Where did you find it? It seems there are quite a few methods to deal with it. $\endgroup$ Commented Sep 17, 2022 at 22:33
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    $\begingroup$ @C-RAM It's problem 1.2.22 of the relatively new book by Douglas West, Combinatorial Mathematics. $\endgroup$
    – Addem
    Commented Sep 17, 2022 at 22:42
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    $\begingroup$ Although I should probably mention that in his statement of the problem he credits Dilcher and Woord for two different publications containing the equality. $\endgroup$
    – Addem
    Commented Sep 17, 2022 at 22:45
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    $\begingroup$ A small remark, this is the so-called Dilcher's formula, see e.g. mathworld.wolfram.com/DilchersFormula.html. $\endgroup$
    – Marian G.
    Commented Jan 30, 2023 at 8:59

1 Answer 1

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Let $S_{m,n}$ be your first sum, so \begin{equation} \begin{split} S_{m+1,n}&=\sum_{1\leq k_1\leq\cdots\leq k_{m+1}\leq n}\frac{1}{k_1k_2\cdots k_mk_{m+1}}\\ &=\sum_{\ell=1}^n\left[\sum_{1\leq k_1\leq\cdots \leq k_m\leq\ell}\frac{1}{k_1k_2\cdots k_m\ell}\right]\\ &=\sum_{\ell=1}^n\frac{1}{\ell}S_{m,\ell}\\ \end{split} \end{equation} Now, we proceed by induction on $m$. Suppose that \begin{equation} S_{m,n}=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}{n\choose k} \end{equation} for all $n\geq 0$, then \begin{equation} \begin{split} S_{m+1,n}&=\sum_{\ell=1}^n\frac{1}{\ell}S_{m,\ell}\\ &=\sum_{\ell=1}^n\frac{1}{\ell}\sum_{k=1}^\ell\frac{(-1)^{k-1}}{k^m}{\ell\choose k}\\ &=\sum_{\ell=1}^n\sum_{k=1}^\ell\frac{(-1)^{k-1}}{\ell k^m}{\ell\choose k}\\ \end{split} \end{equation} Changing the order of summation gives us \begin{equation} \begin{split} S_{m+1,n}&=\sum_{k=1}^n\sum_{\ell=k}^n\frac{(-1)^{k-1}}{\ell k^m}{\ell\choose k}\\ &=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}\sum_{\ell=k}^n\frac{1}{\ell}{\ell\choose k}\\ &=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}\sum_{\ell=k}^n\frac{1}{k}{{\ell-1}\choose {k-1}}\\ &=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^{m+1}}\sum_{\ell=k}^n{{\ell-1}\choose {k-1}}\\ \end{split} \end{equation} Finally, using a modified version of the hockey-stick identity $\sum_{\ell=k}^n{{\ell-1}\choose {k-1}}={n\choose k}$ (when $n\geq k$), we have that \begin{equation} S_{m+1,n}=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^{m+1}}{n\choose k}\\ \end{equation} which finishes our inductive step. Now, note that there exists one tuple of length $0$, and the empty product is defined as $1$, so by definition \begin{equation} S_{0,n}=1=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^0}{n\choose k} \end{equation} where the second equality is true by the binomial theorem, which proves the base case. Therefore, by induction \begin{equation} S_{m,n}=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}{n\choose k} \end{equation} for all $m,n\geq 0$.

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