Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.