A criterion for series convergence? I have a conjecture regarding series convergence that feels like it would be a useful tool to me if I could prove it, but I have been unable to prove or disprove it.
Let $\sum a_n$ be a series of nonnegative terms. Define $\lambda(N)$ to be the number of terms of the series that are greater than $1/N$. 
Conjecture: If $\lambda(N)=O(N^\alpha)$, with $0<\alpha<1$, then $\sum a_n$ converges.
The conjecture is based on the fact that this is true for series of the form $\sum 1/n^k$, $k$ constant (let $\alpha = 1/k$), and my intuition that the hypothesis of the conjecture is enough to make $\sum a_n$ "sufficiently similar to or bounded by" such a series.
Can you offer a counterexample or point me toward an idea for a proof?
(I tried to bound $\Delta\lambda(N) = \lambda(N+1)-\lambda(N)$ from the assumption $\lambda(N)=O(N^\alpha)$, since possibly excluding a finite number of terms, the series sum is bounded above by $\sum \Delta\lambda(N)/N$; but I couldn't see how to do this.)

This is summation by blocks in disguise. 
For every $k\ge0$, the sum of the terms $a_n$ such that $2^{-(k+1)}\le a_n\le 2^{-k}$ is at most the number of terms times a common upper bounds of these, hence at most $\lambda(2^{k+1})$ times $2^{-k}$. It happens that the number $\lambda(1)$ of terms $a_n$ such that $a_n\ge1$ is finite (1), hence it suffices to consider the sum of the terms  $a_n$ such that $a_n\le1$. This sum is at most
$$
\sum_{k=0}^{+\infty}\sum_na_n\cdot[2^{-(k+1)}\le a_n\le 2^{-k}]\le\sum_{k=0}^{+\infty}\lambda(2^{k+1})2^{-k}.
$$
Now, if $\lambda(N)\le c\cdot N^a$ for every $N$ large enough, then $\lambda(2^{k+1})\le c\cdot2^{a(k+1)}$ for every $k$ large enough and the last series above is controlled by the series
$$
\displaystyle2^ac\cdot\sum_k2^{-(1-a)k},
$$ 
which converges since the hypothesis that $a<1$ implies that $2^{-(1-a)}<1$.
(1) The proof is simple: $\lambda(1)\le\lambda(N)$ for every $N\ge1$ and the hypothesis that $\lambda(N)=O(N^a)$ implies that $\lambda(N)$ is finite for $N$ large enough.
