Let's consider a monotone increasing sequence $(K_N) \subseteq \mathbb{N}$ with $(K_N) \xrightarrow[N]{} \infty$ and $(K_N) = \mathrm o(N)$ (less increasing than $(N)$).
Question: $\sum_{j=K_N + 1}^N 1/j \longrightarrow_N 0\,?$
Question: What are necessary conditions for $(a_j) \subseteq \mathbb{R}$ such that $\sum_{j=K_N + 1}^N|a_j|\longrightarrow_N 0\,?$
To 1. Question: I think it doesn't work, but I'm not sure: I tried to prove this with the Euler's constant $\gamma$ - if $\sum_{j=1}^{K_N}1/j - \text{ln}(K_N) \longrightarrow_N \gamma$ (does the limit depend on the asymptotic behaviour of $(K_N)$ ? ) it shouldn't work in general (e.g. set $(K_N) = (\lfloor N^{1/2}\rfloor)$).
To 2. Question: I could imagine that it works for absolutely summable $(a_j) \in \ell^1.$
Thanks!
Consider $K_N=\lfloor \sqrt N\rfloor$. Then $K_N/N\to0$, and $$ \sum_{j=K_N}^N\frac1j\geq\sum_{j=K_N}^N\frac1N=\frac{N-\lfloor \sqrt N\rfloor+1}N\xrightarrow[N\to\infty]{}1. $$
As you say, if $a\in\ell^1(\mathbb N)$, then $$ \sum_{j=K_N + 1}^N|a_j|\leq\sum_{j=K_N + 1}^\infty|a_j|\longrightarrow_N 0. $$ So $a\in\ell^1 (\mathbb N) $ is a sufficient condition. An obvious necessary condition is $a_n\to0$. It might be possible to find a stronger necessary condition, but I'm not sure.