# Order of magnitude and absolute convergence

In connection with studying absolute convergence of Fourier series, I started to wonder about the following:

Assume that $$S\sim \sum_0^{\infty}a_n,$$ where all we know about $a_n$ is that $a_n = o(n^{-1})$ for large values of n, that is, $n\cdot a_n \rightarrow 0$ as $n\rightarrow \infty$. Can we from this conclude that $S$ converges absolutely?

My gut feeling is that this is not enough to guarantee that the series converges absolutely, but I don't know of any counter-examples. I read volume 1, chapter 6, theorem (3.8) in Zygmund's book on trigonometric series, which says that if a function is absolutely continuous and it's derivative is in $L^p$ for some $p>1$, then it's Fourier series converges absolutely. Why does the theorem not include the case $p=1$ if I'm wrong about all this?

## 1 Answer

Let $a_{n}=\frac{1}{n\log(n+1)}$. This should be a sufficient counter-example, since $\sum a_n$ does not converge.

• Is this a well-known result? I remember seeing it somewhere before, but I don't know how to show that this series diverges. – Scounged May 22 '16 at 21:40
• A good proof is in Rudin's PMA – ervx May 22 '16 at 21:41
• Nice, thanks. I'll go look it up. – Scounged May 22 '16 at 21:42
• Theorem 3.29 (bottom of page 62). It says the series $\sum_{2}^{\infty}\frac{1}{n(\log n)^{p}}$ converges iff $p>1$. – ervx May 22 '16 at 21:43