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Let $S_n:= \displaystyle\sum_{k=1}^n \displaystyle\sum_{j=1}^{k-1}\dfrac{(-1)^k}{[j(k-j)]^p} $. Determine whether $\lim\limits_{n\to\infty} S_n =\displaystyle\sum_{k=1}^\infty \displaystyle\sum_{j=1}^{k-1}\dfrac{(-1)^k}{[j(k-j)]^p}$ converges or diverges for $\frac{1}2 < p \leq 1.$

I claim that $S_n$ converges as $n\to \infty.$ Let $c_k:= \displaystyle\sum_{j=1}^{k-1}\dfrac{1}{[j(k-j)]^p}>0\,\forall k\in\mathbb{N},$ with $c_1 := 0.$ We want to show that $c_k$ is eventually nonincreasing (ie. $\exists N\in\mathbb{N}$ such that $\forall k\geq N, c_{k+1}\geq c_k$) and that $\lim\limits_{k\to\infty}c_k=0,$ which will show by Leibnitz's alternating series test that the series converges.

However, this seems very difficult to do.

I want to avoid using "well-known" theorems, and instead use more fundamental theorems like Leibnitz's alternating series test, the Cauchy product formula for absolutely convergent series, etc.

  • the series looks convergent to me as the $k$ term looks to be roughly $c_p\frac{1}{k^{2p-1}}, p \ne 1$ and $c\frac{\log k}{k}, p=1$ so when you alternate it, it converges – Conrad Apr 03 '20 at 02:49
  • @Conrad so how would you prove that claim? –  Apr 03 '20 at 02:52
  • integral estimates for $(\frac{1}{x}+\frac{1}{k-x})^p$ as the $k$ term is the sum of terms of $\frac{1}{k^p}(\frac{1}{j}+\frac{1}{k-j})^p$; they do not seem that straightforward when $p \ne 1$ though, so will have to think more, but I still think the estimate holds; $p=1$ is immediate though; another idea is to show directly that $c_k - c_{k+1}=d_k-d_{k+1}+a_k, d_k \ge d_{k+1}, \sum {|a_k|}$ convergent – Conrad Apr 03 '20 at 03:22

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As already observed, the sum, in fact, converges.

The proof of "eventual monotonicity" can be avoided using a simple trick. We have $$\sum_{k=2}^{n}(-1)^k\sum_{j=1}^{k-1}a_j a_{k-j}=\left(\sum_{k=1}^{n}(-1)^k a_k\right)^2-\sum_{k=n+1}^{2n}(-1)^k\underbrace{\sum_{j=k-n}^{n}a_j a_{k-j}}_{=A_{n,k}}.$$ Now if $0\leqslant a_{k+1}\leqslant a_k$ for all $k$, then $\color{blue}{0\leqslant A_{n,k+1}\leqslant A_{n,k}}$ for $n<k<2n$, which gives $$\left|\sum_{k=2}^{n}(-1)^k\sum_{j=1}^{k-1}a_j a_{k-j}-\left(\sum_{k=1}^{n}(-1)^k a_k\right)^2\right|\leqslant A_{n,n+1}=\sum_{k=1}^{n}a_k a_{n+1-k}.$$ This applies to our $a_k=k^{-p}$, and it remains to prove $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n-1}\big(k(n-k)\big)^{-p}=0$ for $1/2<p\leqslant 1$.

For me, the easiest (if not elementary) way for $p\neq 1$ is, indeed, to use an integral estimate: $$\sum_{k=1}^{n-1}\big(k(n-k)\big)^{-p}=n^{-2p}\sum_{k=1}^{n-1}\left[\frac{k}{n}\left(1-\frac{k}{n}\right)\right]^{-p}\leqslant n^{1-2p}\int_0^1\big(x(1-x)\big)^{-p}dx$$ (look carefully at the "lower Darboux sum" for the integrand); alternatively, one could start with $\require{action}\texttip{\color{blue}{(x+y)^p\leqslant x^p+y^p}}{taken at $x=1/k$ and $y=1/(n-k)$}$ and prove that $\sum_{k=1}^{n-1}k^{-p}=\mathcal{O}(n^{1-p})$, but I think it's a longer road.

For $p=1$, one could simply check that $n\mapsto\sum_{k=1}^{n-1}\frac{1}{k(n-k)}=\frac2n\sum_{k=1}^{n-1}\frac1k$ is decreasing.

metamorphy
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