Convergence of alternating harmonic sums - MathOverflow

Convergence of alternating harmonic sums

2011-01-13T02:01:55Z

I owe the idea of asking this question to Max Muller and his curiosity.

What is the set of $\alpha$ in the interval $0\le\alpha < 1$ for which the alternating sum $$ \sum_{n=1}^\infty\frac{(-1)^{n+[n^\alpha]}}n $$ converges? Here $[\ \cdot\ ]$ denotes the integral part of a number.

It clearly converges when $\alpha=0$ and my post to Max's answer implies the convergence when $\alpha=1/2$. What about more general $\alpha$? Of course, the question is meaningful for positive $\alpha\notin \mathbb Z$ as well, but then it seems to be much harder.

Answer by Leandro for Convergence of alternating harmonic sums

2011-01-13T06:16:58Z

This is not an answer, but it is too long for a comment.

Hi Wadim, nice problem. I was trying to obtain a partial answer for it based on the following

Proposition. Let be $\xi_1,\xi_2,\ldots$ a sequence of independent Bernoulli random variables with $\mathbb{P}(\xi_n=+1)=\mathbb{P}(\xi_n=-1)=\frac{1}{2}$, then the series $\sum \xi_n a_n$, with $|a_n|\leq c$, converges with probability 1, if and only if $\sum a_n^2<\infty$.

The idea it was consider $a_n=\frac{1}{n}$, and try to analyse the set $$ A:=\left\{\Big((-1)^{n+[n^{\alpha}]}\Big)_{n\in\mathbb{N}}; 0\leq \alpha<1 \right\} \subset \{-1,1\}^{\mathbb{N}}. $$
In case that $\mathbb{P}(A)\neq 0$, since this product measure has no atoms we could, at least, say that the set of $\alpha$'s for which the series is convergent is non-enumerable.

Answer by Anthony Quas for Convergence of alternating harmonic sums

2011-01-13T06:31:33Z

It converges for all $0\le\alpha<1$. Define the $k$-block to be the set of $n$ such that $[n^\alpha]=k$ (it ranges from $\lceil k^{1/\alpha}\rceil$ to $\lceil (k+1)^{1/\alpha}\rceil-1$).

The absolute value of the contribution to the sum from the $k$-block is at most the reciprocal of its left endpoint (the terms form an alternating series) - that is approximately $k^{-1/\alpha}$. Hence the contributions from the $k$-blocks are absolutely summable. Bingo

Answer by Anthony Quas for Convergence of alternating harmonic sums

2011-01-13T07:54:42Z

Let me try and give an answer for $\alpha>1$ also. This one uses some technology from a paper of mine with Boshernitzan, Kolesnik and Wierdl.

I want to use exponential sums. Using the notation of that paper we take $a(n)=n+n^\alpha$. That paper lets us control $\hat A_t(1/2)=(1/t)\sum_{n\le t}e([a(n)]/2)=(1/t)\sum_{n\le t}(-1)^{n+[n^\alpha]}$ where $e(x)=e^{2\pi i x}$ (see after Lemma 7.2).

The proofs of Theorem 3.4 and Theorem 7.1 give (if you read carefully) the existence of an $\epsilon>0$ and a $C$ such that $|\hat A_t(1/2)| < Ct^{-\epsilon}$ for all $t$. This says that the difference between the number of $+1$'s and the number of $-1$'s (the discrepancy) for $n\le t$ is at most $t^{1-\epsilon}$ (ignoring constants from now on).

Now let $K>2/\epsilon$ and divide the integers into blocks $I_j=(j^K,(j+1)^K]$. The discrepancy up to $j^K$ is at most $j^{K-2}$ by the above. Similarly the discrepancy up to $(j+1)^K$ is also at most $j^{K-2}$. So the discrepancy in the $I_j$ block is at most $j^{K-2}$.

We now have $\sum_{n\in I_j}(-1)^{n+[n^\alpha]}/n = \sum_{n\in I_j}(-1)^{n+[n^\alpha]}/j^K + \sum_{n\in I_j}(-1)^{n+[n^\alpha]}(1/n-1/j^K)$.

By the above comment, the first term contributes $j^{-2}$. In the second term, there are $j^{K-1}$ terms, each contributing in absolute value at most $j^{-K-1}$ giving a maximum total contribution of $j^{-2}$.

It follows that the contributions from the $I_j$-blocks are absolutely summable and we're done.

Of course $\alpha<0$ is trivial so this is good for all $\alpha$ except the positive integers.