Another way of thinking about it:
Let $f(n):=a_n$ and extend $\lambda(N)$ to $\mathbb R$ by defining $\lambda(t)=\#\{n:f(n)>1/t\}$ (here $\#$ means cardinality). Define $F_f(t)=\lambda(1/t)=\#\{n:f(n)>t\}$; if we think of the set of positive integers as a measure space, with the counting measure, then the function $F_f(t)$ is the distribution function of $f$, and $ \sum_{n=1}^{\infty}f(n)=\int_0^{\infty}F_f(t)\,\mathrm dt. $ Your hypothesis is that $F_f(1/N)=O(N^{\alpha})$. In particular, there are only finitely many values of $f$ that are $>1$, and $f$ is bounded.
Since $\lceil t^{-1}\rceil^{-1}\leq t<2\lceil t^{-1}\rceil^{-1}$ for $t\leq 1$, we can convert the bound on $F_f(1/N)$ to a bound on $F_f(t)$: $F_f(t)\leq F_f(\lceil t^{-1}\rceil^{-1})\leq C\lceil t^{-1}\rceil^{\alpha}\leq C2^{\alpha} t^{-\alpha}\qquad(0 for some constant $C$. That is to say, $F_f(t)=O(t^{-\alpha})$. (This is the same as saying that $f$ is in weak $\ell^{\alpha}$.) Now we can just approximate the integral above: If $M$ is a bound for $f$, then \int_0^{\infty}F_f(t)\,\mathrm dt = \int_0^MF_f(t)\,\mathrm dt \leq C'\int_0^1t^{-\alpha}\,\mathrm dt + C'MF_f(1) <\infty, since $0<\alpha<1$.
In fact, your observation can be generalized: Let $L^{p,\infty}(\mu)$ denote weak $L^p(\mu)$, the space of all functions $f:(X,\mu)\to\mathbb C$ satisfying $ \|f\|_{p,w}:=\sup_{t>0}\,\,{t\,\mu(\{x:|f(x)|>t\})^{1/p}}<\infty. $ If $f\in L^{p,\infty}(\mu)\cap L^{q,\infty}(\mu)$, where $0, then $f\in L^r(\mu)$ for all $r\in (p,q)$. (See Grafakos, Classical Fourier Analysis, pp. 8-9 for a more precise result; also, Terry Tao's blog has some interesting information in connection with this topic in general.)
To apply this generalization to your specific case, you can take $p=\alpha$, $q=\infty$ (because $\{a_n\}$ is bounded), and $r=1$ to conclude that $\{a_n\}\in \ell^1$.
(I wanted to say something about this yesterday, but I couldn't think of the appropriate generalization. Being new to the subject, I also wanted to check with a professor of mine to ensure that the above analogy was apt. Please let me know if there are errors.)