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Here's the setup: let $N(m)$, $m = 1,2, \cdots$ be a sequence of nonnegative integers for which

$(*)~~~~~\sum_{m = 1}^{\infty} \frac{N(m)}{m} < \infty$

Now let $C>0$ be any constant. My question is, does there exist a constant $K < \infty$ such that for all $n \geq 1$,

$\left(1+C \sum_{m = n+1}^{\infty} \frac{N(m)}{m}\right)^{\sum_{m = 1}^n N(m)} < K$

I already know the bound $\sum_{m = 1}^n N(m) < n$ for $n$ sufficiently large (uses Abel summation), but for some reason I can't seem to make the above work. For instance, one can check that

$\lim\limits_{n \rightarrow \infty} (1 + \frac{1}{\log n})^n = \infty$

but I'm not sure if this is a death knell for what I want to show. My confusion stems from the idea that the condition $(*)$ seems to suggest that "most" of the $N(m)$ are $0$, hence $\sum_{m = 1}^n N(m)$ ought to be much, much smaller than $n$.

Can anybody think of a counterexample?

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Where does the question arise? –  Yemon Choi Sep 27 '12 at 23:20
    
I am very sleep-deprived right now but one idea that comes to mind is to fix an infinite subset S of the natural numbers and take $N(\cdot)$ to be the indicator function/characteristic function of S. If you take S to be the "nearest integer perturbation" of the set $\{ k^{3/2} : k \in {\bf N}\}$ then some back of envelope calculations suggest $\sum_{m=1}^n N(m)$ is approx $n^{2/3}$ while $\sum_{m=n+1}^\infty N(m)/m$ is $O(n^{-1/3})$. Which looks like it would give a counter-example... however, this is all being done through a mental fog & I may have made a stupid error –  Yemon Choi Sep 27 '12 at 23:31

1 Answer 1

up vote 1 down vote accepted

Counterexample: Define $N(m)$ by $$N(m):=\begin{cases} \lfloor n!/2\rfloor &\text{if } m=(n+2)!-1\\\\ \lceil n!/2\rceil &\text{if } m=(n+2)!\\\\ 0 &\text{otherwise}\\\\ \end{cases}$$ and note that $\sum\limits_{m=1}^\infty \frac{N(m)}{m}<\sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$, yet $$\begin{align} \left(1+\sum\limits_{m=(n+2)!}^{\infty} \frac{N(m)}{m}\right)^{\sum\limits_{m=1}^{(n+2)!-1} N(m)} &\ge \left(1+\frac{1}{2(n+1)(n+2)}\right)^{n!/2}\\\\ &\ge 1+\frac{n!/2}{2(n+1)(n+2)}\to \infty\\\\ \end{align}$$ thus clearly no upper bound can exist.

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Nice - looks similar to the construction I outlined. –  Yemon Choi Sep 28 '12 at 4:52
1  
Yes, I think basically any function which has sparse, sharp peaks which come close (in the sense of $1/n^{1+\epsilon}$) to their index will provide a counterexample. –  Alex Becker Sep 28 '12 at 5:12
    
Yep, that's a counterexample. Oh well; this came up when I was trying to find a bound on the total variation of a sequence of signed measures. I think this is true if $\sum_{m = 1}^n N(m) \leq C \sqrt{n}$. Then the tail of the sum looks like $1/\sqrt{n}$ by using Abel summation. Do you think any weaker bound (e.g. $\sum_{m = 1}^n N(m) = O(n^{1-\epsilon}$ for $\epsilon < 1/2$) would work? –  A Blumenthal Sep 28 '12 at 21:38
1  
Sorry, Yemon Choi's earlier reply answers my question in the negative. –  A Blumenthal Sep 28 '12 at 21:39

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