Question

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?

Answer 1

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.