# Convergence of alternating harmonic sums

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.

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I'd like to add a close question. Do you know non-periodic sequences $(\epsilon_n)$ of $\pm1$ (for $n\ge1$) that give a convergent series $\sum_n \epsilon_n/n$ that can be computed? Periodic sequences lead to an integral of a rational function easily computed. Max Muller series gave an integral of a theta function, that however didn't seem to be related to anything known. –  Pietro Majer Jan 13 '11 at 8:09
@Pietro, I'd like your variation (where and when was it closed?). It looks like the only reasonable cases corresponding to the integrals of non-rational functions are... theta functions. In other cases, the functions we integrate have practically no chance to be "special" (=known). The integrals of weight 1/2 modular forms (our thetas) are not well understood but there is some literature about them. In any case, they are integrals of special functions! –  Wadim Zudilin Jan 13 '11 at 9:43
@Pietro : To answer your question, I can only give the following artificial example. Start from the convergent series $\sum_{n \geq 1} (-1)^n/n$. Then for any even $n$ change the $+1$ to $-1$ whenever $n$ is a square. Then the resulting series differs from the original one by $2 \sum_{n \geq 1} \frac{1}{(2n)^2} = \frac{\pi^2}{12}$. –  François Brunault Jan 13 '11 at 12:24
@pietro I wonder about the sums $\sum_{n=1}^\infty\frac{(-1)^{n+[n\alpha]}}n$ and $\sum_{n=1}^\infty\frac{(-1)^{[n\alpha]}}n$. Maybe for some $\alpha$ such as $1+\sqrt{2}$ or $\frac{1+\sqrt{5}}{2}$ (or maybe their reciprocals would be easier) the sum can be found. A more frivolous comment is that one could implicitly define a non-periodic sequence by starting with the limit moving towards it more or less deliberately. –  Aaron Meyerowitz Jan 13 '11 at 17:02
@Wadim, I think Pietro meant "close" as in "closely related", not as in "closed". –  JBL Jan 13 '11 at 17:41

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

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A nice answer, Anthony. Can't find any bugs. :-) –  Wadim Zudilin Jan 13 '11 at 6:44
I am now confident in that the problem could be posted in class (of course, as advanced problem!). I also see why $\alpha>1$ is harder. Maybe, some time later I'll ask the remaining piece... –  Wadim Zudilin Jan 13 '11 at 6:56

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.

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Thanks, Antony! I'll follow your proof and the related link with care. The idea is however clear to me. –  Wadim Zudilin Jan 13 '11 at 8:05

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.

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Thanks, Leandro, this is an interesting approach. Does it imply that, under the assumption that $[n^\alpha]$ are equidistributed moodulo 2 for a certain set $\alpha\in A$, that the series converges for almost all $\alpha\in A$? Probably, a little bit more is needed. –  Wadim Zudilin Jan 13 '11 at 6:25
In fact the set $A$ is a subset of the set of sequences where the density of occurrences of two consecutive $+1$'s is 0. This set has probability 0. –  Anthony Quas Jan 13 '11 at 8:31