Show that $\limsup_{k \to \infty} 2^{-k} N_k = 0$ where $N_k$ is the number of $a_n \geq 2^{-k}$. Let $\{a_n\}_{n \geq 1}$ be a non-negative sequence of reals such that $\sum_{n \geq 1} a_n$ converges to $s$. Define $N_k = |\{n \in \mathbb{N} : a_n \geq 2^{-k}\}|$. Show that 
\begin{equation} \limsup_{k \to \infty} 2^{-k} N_k = 0 \end{equation}
The idea I had was to bound the terms $v_N :=\sup\{2^{-k} N_k : k \geq N\}$ by $s c_n$ where $c_n$ is some sequence converging to 0, though this approach didn't seem to go anywhere. 
Some other facts I observed were that for all $k \in \mathbb{N}$ there exists $n_k \in \mathbb{N}$ such that $a_n < 2^{-k}$ for all $n \geq n_k$ (just the definition of convergence), and I was thinking there would be a way to split up the partial sums and write in the form of $2^{-k} N_k + \text{stuff}$, though I couldn't work it out. Any suggestions would be appreciated. 

Let $E_k=\{n: 2^{-(k+1)} \leq a_n < 2^{-k}\}$. Then $\sum_k \sum_{n \in E_k} a_n \leq \sum_n a_n$ because $E_k$ 's are disjoint. Since $a_n \geq 2^{-(k+1)}$ for $n \in E_k$ we get $\sum_k \sum_{n \in E_k} 2^{-(k+1)}  \leq \sum_n a_n$. This gives $\sum_k 2^{-(k+1)} (N_{k+1} -N_k) <\infty$. Use partial summation to conclude that $\sum N_k b_k <\infty$ where $b_k =\sum_{j=1}^{k} 2^{-(j+1)}$. Conclude the proof by computing the geometric sum $\sum_{j=1}^{k} 2^{-(j+1)}$.
