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?