Binomial Coefficient Challenge 8!

Prove -

∑k=0n−1(nk)(−1)n−k−1n−kkn=nn(Hn−1)\displaystyle \sum_{k=0}^{n-1}{\binom{n}{k} \frac{{(-1)}^{n-k-1}}{n-k} k^n} = n^n (H_n -1)

where HnH_n is nnth Harmonic number.

Bonus

∑k=1n(nk)(−1)k−1k(x+ky)n=xn−1(xHn+ny)\displaystyle \sum_{k=1}^n{\binom{n}{k}\frac{{(-1)}^{k-1}}{k} {(x+ky)}^n} = x^{n-1} (x H_n + n y)

Note by Kartik Sharma
3 years, 12 months ago

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Comments

Since ∑k=1n(nk)(−1)k−1xk−1  =  x−1−∑k=0n(nk)(−1)kxk−1  =  1−(1−x)nx \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 xx, we deduce that ∑k=1n(nk)(−1)k−1k−1  =  ∫011−(1−x)nx dx  =  ∫011−xn1−x dx  =  Hn \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≥1n \ge 1.

Next, ∑k=0n(nk)(−1)kkj  =  ∑k=0n(nk)(−1)n−k(n−k)j  =  (−1)nC(n,j) \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,jn,j, where C(n,j)C(n,j) is the number of surjections from a set of size jj to a set of size nn. In particular, then, ∑k=0n(nk)(−1)kkj  =  00≤j<n \sum_{k=0}^n {n \choose k}(-1)^k k^j \; = \; 0 \hspace{2cm} 0 \le j < n Thus ∑k=1n(nk)(−1)k−1k(x+ky)n=∑j=0n(nj)xn−jyj∑k=1n(nk)(−1)k−1kj−1=∑j=01(nj)xn−jyj∑k=1n(nk)(−1)k−1kj−1+∑j=2n(nj)xn−jyj∑k=0n(nk)(−1)k−1kj−1=∑j=01(nj)xn−jyj∑k=1n(nk)(−1)k−1kj−1=xn∑k=1n(nk)(−1)k−1k−1+nxn−1y∑k=1n(nk)(−1)k−1=xnHn+nxn−1y \begin{aligned} \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{aligned} The main result follows by putting x=nx=n and y=−1y=-1.

Mark Hennings - 3 years, 12 months ago

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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).

Ishan Singh - 3 years, 12 months ago

@Mark Hennings @Ishan Singh

Kartik Sharma - 3 years, 12 months ago
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