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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?

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Is there a typo in the displayed equation? Maybe it is $|x_j|$ instead of $|x_n|$ inside the sum. –  Siminore Aug 29 '12 at 14:16
    
Thanks for pointing out –  Vivek Aug 29 '12 at 14:21

1 Answer 1

up vote 3 down vote accepted

Since $x_{n} = \mathcal{O}(n^{-1})$ there is some $M>0$ such that $\displaystyle x_n < \frac{M}{n}$ for all $n\in \mathbb{N}.$ Now pick a fixed $\epsilon>0$ and let $\lambda>1$ be arbitrary for now (we will pick it more specially when we need to).

Each term of the sum has a simple bound: For $j>n$, $\displaystyle |x_j| < \frac{M}{j} < \frac{M}{n}.$ There are $(\lambda -1)n$ terms in the sum, so $$ \sum_{n < j \leq \lambda n} |x_n| < (\lambda-1)n\cdot \frac{M}{n}= (\lambda-1)M.$$

Thus if we pick $\lambda$ so that $(\lambda-1)M<\epsilon$ then $$ \sum_{n < j \leq \lambda n} |x_n| < \epsilon.$$

The result in Katznelson is the result of taking the limit superior the above inequality.

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