I came across this in Katznelson's Harmonic Analysis.
It is claimed that if $\{x_n\}$ is a series such that $x_n=O(\frac{1}{n})$, then for every $\epsilon>0$ there is a $\lambda>1$ such that $ \limsup_{n\to \infty}\sum_{n<|j|\leq \lambda n}|x_n|<\epsilon $
$x_n=O(\frac{1}{n})$ implies that there is an $M$ such that for all large $|n|$, $|x_n|\leq\frac{M}{n}$
Using this I could conclude that if $\epsilon>2M$, then if $\lambda$ is chosen such that $\lambda<\exp(\frac{\epsilon}{2M}-1)$ then $\sum_{n<|j|\leq \lambda n}|x_j|<\epsilon$ for all large $n$. I could do this using the fact that if $H(n)=\sum_{j=1}^{n}\frac{1}{j}$, then we have $\log(n+1)\leq H(n)\leq 1+\log(n)$.
Could someone give me a hint for the other case. Can the same approach be modified to obtain the result or will I have to get involved in the nitty-gritty of limsup?