Find $\sum^{n}_{k=1} \binom{n}{k} \frac{(-1)^k}{k^a}$

Question

I have been trying to find a general representation of the following finite sum.

$$ S(n,a) = -\sum^{n}_{k=1} \binom{n}{k} \frac{(-1)^k}{k^a} $$

The sum seems to be related to Generalized Harmonic Numbers, as I've been able to work out the first few integer values of $a$ as the following.

$$ S(n,1) = H_{n} \ , \ S(n,2) = \frac{(H_{n})^{2} + H_{n}^{(2)}}{2} \ , \ S(n,3) = \frac{(H_{n})^3}{6} + \frac{H_{n}^{(2)}H_{n}}{2} + \frac{H_{n}^{(3)}}{3} $$

Where $ H_{n}^{(m)}$ is the n-th m-th order generalized harmonic number. It also sees to be closely related to the Stirling numbers of the second kind because if you take $a$ to be negative you get something very close to the explicit definition of the stirling numbers. I have looked online to see if there is any identities related to these special numbers and function that involve a sum like this but I haven't been able to find anything. If you have any insight please let me know.

Answer 1

Recall the complete symmetric function in $n$ variables: $$ h_{n,k}:=h_{n,k}(x_1,\dots,x_n)=\sum_{1\le i_1\le \dots \le i_k \le n} x_{i_1}x_{i_2}\cdots x_{i_k} $$ Note that this satisfies the recurrence $$ \begin{align} h_{n,k}&=h_{n-1,k}+x_n\cdot h_{n,k-1}\tag1 \\ h_{n,0}&=1 \quad (n>0) \\ h_{0,k}&=0 \quad (k>0) \end{align} $$ The recurrence in $(1)$ uniquely defines the bivariate sequence $h_{n,k}$.

Claim: For all $n,a\in \mathbb N$ where $n>0$ or $a>0$, $$ S(n,a)=h_{n,a}(1,\tfrac12,\tfrac13,\dots,\tfrac1n) $$

Proof: To prove this, note that $$ \begin{align} S(n,a) &= \sum_{i=1}^n(-1)^{i-1}\binom {n-1}{i}i^{-a}&+&&&\sum_{i=1}^{n}(-1)^{i-1}\binom{n-1}{i-1}i^{-a} \\ &= S(n-1,a)&+&&&\frac1n\sum_{i=1}^n(-1)^{i-1}\binom {n-1}{i-1}\cdot \frac{n}{i}\cdot i^{-(a-1)} \\ &= S(n-1,a)&+&&&\frac1n\sum_{i=1}^n(-1)^{i-1}\binom {n}{i}\cdot i^{-(a-1)} \\ &=S(n-1,a)&+&&&\tfrac 1nS(n, a-1) \end{align} $$ Therefore, letting $x_n=\frac1n$ for each $n\ge 1$, we see $S(n,a)$ satisfies the exact same recurrence and base cases as $(1)$. This gives a proof that $S(n,a)=h_{n,a}(1^{-1},2^{-2},\dots,n^{-1})$ by induction on $n+a$. $\tag*{$\square$}$

Using Wikipedia, we can express the homogenous symmetric polynomials in terms of the power sum polynomials $p_{n,k}$, which evaluated at $1,\frac12,\dots,\frac1n$ is just the generalized Harmonic numbers, $H_{n,k}$. $$ S(n,a)=h_{n,a}= \sum_{\lambda |- a} P(\text{random $\pi\in S_a$ has cycle type $\lambda$}) \prod_{i=1}^{\text{len}(\lambda)}H_{n, \lambda_i} $$ For example, $$ \begin{align} S(n,4)&=\frac1{4!}\big(\quad 1\cdot (H_{n,1})^4 && \text{1 permutation with cycle type $(1,1,1,1)$} \\ &\hspace{1.3cm} + 6\cdot (H_{n,1})^2\cdot H_{n,2} && \text{6 permutations with cycle type $(2,1,1)$} \\ &\hspace{1.3cm} + 3\cdot (H_{n,2})^2 && \text{3 permutations with cycle type $(2,2)$} \\ &\hspace{1.3cm} + 8\cdot H_{n,1}\cdot H_{n,3} && \text{8 permutations with cycle type $(3,1)$} \\ &\hspace{1.3cm} + 6\cdot H_{n,4}\big) && \text{6 permutations with cycle type $(4)$} \end{align} $$ As $n\to\infty$, the dominant term is always the identity permutation, so we get $S(n,a)\sim \frac{(\log n)^a}{a!}$ as $n\to\infty$.

Answer 2

The following solution is already given in this post but I am providing more details here: $$\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k^{a}}=\sum_{k=1}^n \binom{n}{k}(-1)^{k-1}\int_0^1 \frac{(-1)^{a-1}}{(a-1)!} x^{k-1}\ln^{a-1}(x)dx$$ $$=\frac{(-1)^{a-1}}{(a-1)!}\int_0^1\left(\sum_{k=1}^n \binom{n}{k}(-x)^{k-1}\right)\ln^{a-1}(x)dx$$ $$=\frac{(-1)^{a-1}}{(a-1)!}\int_0^1\left(\frac{1-(1-x)^n}{x}\right)\ln^{a-1}(x)dx$$ $$\overset{1-x\to x}{=}\frac{(-1)^{a-1}}{(a-1)!}\int_0^1\frac{1-x^n}{1-x}\ln^{a-1}(1-x)dx$$ $$\overset{IBP}{=}\frac{(-1)^a n}{a!}\int_0^1 x^{n-1}\ln^a(1-x)dx$$ $$=\frac{(-1)^a n}{a!}\lim_{m\to1}\frac{\partial^a}{\partial m^a}\operatorname{B}(m,n)$$ $$=\frac{(-1)^a n}{a!} t(a,n).$$ Let's find the derivative of the beta function, $$ \frac{\partial}{\partial m}\operatorname{B}(m,n)=\frac{\partial}{\partial m}\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$ $$=\Gamma(m)\frac{\Gamma(m+n)\Gamma(n)\psi(n)-\Gamma(m+n)\psi(m+n)\Gamma(n)}{\Gamma^2(m+n)}$$ $$=-\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}(\psi(m+n)-\psi(n))$$ $$=-\operatorname{B}(m,n)(\psi(m+n)-\psi(n))=-\sum_{k=0}^{n-1}\frac{\operatorname{B}(m,n)}{k+m}.$$ Taking the $a$-th derivative, we find $$\frac{\partial^a}{{\partial m}^{a}}\operatorname{B}(m,n)={(-1)}^a(a-1)!\left(\sum_{k=0}^{n-1}\frac{\operatorname{B}(m,n)}{(k+m)^{a}}+\sum_{j=1}^{a-1}\frac{{(-1)}^{a-j}}{j!}\sum_{k=0}^{n-1} \frac{\frac{\partial^j}{{\partial m}^{j}}\operatorname{B}(m,n)}{{(k+m)}^{a-j}}\right).$$ Let $m\to1$ and remember $\displaystyle\lim_{m\to1}\frac{\partial^a}{{\partial m}^{a}}\operatorname{B}(m,n)=t(a,n)$ observing that $$\lim_{m\to1}\sum_{k=0}^{n-1} \frac{1}{{(k+m)}^r}=H_n^{(r)};\quad \lim_{m\to1}\operatorname{B}(m,n)=\frac{1}{n},$$ we get $$t(a,n)={(-1)}^a(a-1)!\left(\frac{H_n^{(a)}}{n}+\sum_{j=1}^{a-1}\frac{{(-1)}^{j}}{j!}H_n^{(a-j)}t(j,n)\right).$$

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Examples:

For $a=1$,

$$\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k}=-nt(1,n)=H_n$$

For $a=2$,

$$\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k^2}=\frac{n}2 t(2,n)=\frac{1}2\left(H_n^2+H_n^{(2)}\right)$$

For $a=3$,

$$\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k^3}=-\frac{n}6t(3,n)=\frac{1}6\left(H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

For $a=4$,

$$\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k^4}=\frac{n}{24} t(4,n)=\frac{1}{24}\left(H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3(H_n^{(2)})^2+6H_n^{(4)}\right)$$