Prove -

\[\displaystyle \sum_{k=0}^{n-1}{\binom{n}{k} \frac{{(-1)}^{n-k-1}}{n-k} k^n} = n^n (H_n -1)\]

where \(H_n\) is \(n\)th Harmonic number.

**Bonus**

\[\displaystyle \sum_{k=1}^n{\binom{n}{k}\frac{{(-1)}^{k-1}}{k} {(x+ky)}^n} = x^{n-1} (x H_n + n y)\]

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## Comments

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TopNewestSince \[ \sum_{k=1}^n {n \choose k} (-1)^{k-1} x^{k-1} \; = \; x^{-1} - \sum_{k=0}^n {n \choose k} (-1)^k x^{k-1} \; =\; \frac{1 - (1-x)^n}{x} \] for all \(x\), we deduce that \[ \sum_{k=1}^n {n \choose k}(-1)^{k-1} k^{-1} \; = \; \int_0^1 \frac{1 - (1-x)^n}{x}\,dx \; = \; \int_0^1 \frac{1 - x^n}{1-x}\,dx \; = \; H_n \] for all integers \(n \ge 1\).

Next, \[ \sum_{k=0}^n {n \choose k}(-1)^k k^j \; = \; \sum_{k=0}^n {n \choose k}(-1)^{n-k} (n-k)^j \; = \; (-1)^n C(n,j) \] for non-negative integers \(n,j\), where \(C(n,j)\) is the number of surjections from a set of size \(j\) to a set of size \(n\). In particular, then, \[ \sum_{k=0}^n {n \choose k}(-1)^k k^j \; = \; 0 \hspace{2cm} 0 \le j < n \] Thus \[ \begin{align} \sum_{k=1}^n {n \choose k}\frac{(-1)^{k-1}}{k}(x+ky)^n & = \sum_{j=0}^n {n \choose j} x^{n-j} y^j \sum_{k=1}^n{n \choose k} (-1)^{k-1} k^{j-1} \\ & = \sum_{j=0}^1 {n \choose j}x^{n-j}y^j \sum_{k=1}^n {n \choose k}(-1)^{k-1}k^{j-1} + \sum_{j=2}^n {n \choose j}x^{n-j}y^j \sum_{k=0}^n {n \choose k}(-1)^{k-1} k^{j-1} \\ & = \sum_{j=0}^1 {n \choose j}x^{n-j}y^j \sum_{k=1}^n {n \choose k}(-1)^{k-1}k^{j-1} \\ & = x^n \sum_{k=1}^n {n \choose k}(-1)^{k-1} k^{-1} + nx^{n-1}y \sum_{k=1}^n {n \choose k}(-1)^{k-1} \\ & = x^nH_n + nx^{n-1}y \end{align} \] The main result follows by putting \(x=n\) and \(y=-1\).

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Same method. We can also derive the main identity using forward difference operator for example here. Yet to find a method using only elementary summations though (for the main identity).

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@Mark Hennings @Ishan Singh

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