Is $\lim S_{n,m}=\sum_{k=1}^n({-1})^k{n\choose k}k^{-m}<\infty $ for $ n \to \infty$ and $m$ large?

Question

Let $m$ be a fixed positive integer ($m>1)$ and let $$S_{n,m}=\sum_{k=1}^n({-1})^k{n\choose k}k^{-m}$$ be a partial sum of real series.

My question here is : Is $\lim S_{n,m} <\infty $ as $ n \to \infty$?

Thank you for any help.

Answer 1

The approach to the asymptotics, as $n \to \infty$, of sums of the form $$ S_n:=\sum_{k=1}^n({-1})^k{n\choose k}f_k \tag1 $$ is known as the technique of Rice's integrals. The idea is to represent the sum as an integral over an appropriate complex contour then evaluating it by the residue calculus. The residue computation applied here reduces to the following coefficient extraction: $$ \sum_{k=1}^n({-1})^k{n\choose k}\frac1{k^m}=-[s^m]\frac1{\left(1-\dfrac{s}1 \right)\left(1-\dfrac{s}2 \right)\cdots \left(1-\dfrac{s}n \right)} \tag2 $$ and we get, for $m\geq2$, as $n \to \infty$,

$$ -\sum_{k=1}^n({-1})^k{n\choose k}\frac1{k^m}=P_m(\log n)+\mathcal{O}\left(\frac{(\log n)^m}{n}\right) \tag3 $$

where $P_m$ is a polynomial of degree $m$. The above result is proved by Ph. Flajolet and R. Sedgewick in this remarkable paper (pp. 6-7, 1995) where an explicit form of $P_m$ is given.

For example, we have, as $n \to \infty$,

$$ \begin{align} -\sum_{k=1}^n({-1})^k{n\choose k}\frac1{k^2}=\frac12 (\log n)^2+\gamma \log n+C_2+\mathcal{O}\left(\frac{(\log n)^2}{n}\right) \tag4 \end{align} $$

or

$$ \begin{align} &-\sum_{k=1}^n({-1})^k{n\choose k}\frac1{k^3}\\&=\frac16 (\log n)^3+\frac{\gamma}2 (\log n)^2+\left(\frac{\gamma^2}2+\frac{\pi^2}{12}\right) \log n+C_3+\mathcal{O}\left(\frac{(\log n)^3}{n}\right) \tag5 \end{align} $$

where $C_1,C_2$ are constants and $\gamma$ denotes the Euler-Mascheroni constant.

Answer 2

I can show that if $m \ge n-3$, then terms of the sum decrease in absolute value, so the sum of the series is between the first two partial sums, i.e., $-n$ and $-n+\frac{n(n-1)}{2\ 2^m} =-n(1-\frac{n-1}{2^{m+1}}) $.

$S_{n,m} =\sum_{k=1}^n({-1})^k{n\choose k}k^{-m} $

If $r_k ={n\choose k}k^{-m} $,

$\begin{array}\\ s_k &=\frac{r_{k+1}}{r_k}\\ &=\frac{{n\choose k+1}(k+1)^{-m}}{{n\choose k}k^{-m}}\\ &=\frac{\frac{n!}{(k+1)!(n-k-1)!}}{\frac{n!}{k!(n-k)!}}\frac{k^m}{(k+1)^m}\\ &=\frac{n-k}{k+1}\frac{1}{(1+1/k)^m}\\ \end{array} $

If $s_k < 1$, then the series is alternating with decreasing terms, so is bounded between consecutive partial sums.

For $k \ge n/2$, $s_k < 1$.

Also, since $(1+1/k)^m > 1+m/k $, $s_k <\frac{n-k}{k+1}\frac{1}{1+m/k} =\frac{n-k}{k+1}\frac{k}{k+m} $, so $s_k < 1$ if $k(n-k) <(k+1)(k+m) $ or $kn-k^2 <k^2+(m+1)k+m $ or $2k^2+(m+1-n)k+m > 0 $ or $k(2k+m+1-n)+m > 0 $.

Therefore if $m \ge n-3$, $s_k < 1$, so the terms in $S_{m, n}$ decrease in absolute value. Therefore the sum is between any two consecutive partial sums, so $S_{n, m}$ is certainly bounded.