If $\sum a_n$ converges , then $a_n<1/n$ a.e? If $\sum_{n=1}^{\infty}a_n<\infty$ is a Positive convergent series, does the following limit hold?
$$\lim_{n\to\infty}\frac{\mathrm{Card}\{1\leq k\leq n , a_k\geq\frac{1}{k}\}}{n}=0$$
I know of a non-trival example to support this:
$$a_n=1/n ,\mathrm{if} \sqrt{n}\in\mathbb{N},a_n=1/n^2. $$For those who don't know, the Card of $\{1\leq k\leq n , a_k\geq\frac{1}{k}\}$ is $O(\sqrt{n}).$
 A: Note: as noted in the results, the above does not exactly prove what the OP asked.
Indeed, the claimed result holds:

Proposition. Assume that $(a_n)_{n\in\mathbb{N}}$ is a non-negative sequence such that $\sum_{n=1}^\infty  a_n < \infty$. Then, we have
  $$
 \lim\!\inf_{n\to\infty} \frac{|\{ 1\leq k\leq n : a_k \geq 1/k \}|}{n} = 0\,.
$$

Proof. 
Let, for $n\geq 1$, $$S_n \stackrel{\rm def}{=} \{ 1\leq k\leq n : a_k \geq 1/k \} \tag{1}$$
and assume by contradiction that
$$
\lim\inf_n\frac{|S_n|}{n}=\alpha > 0\,. \tag{2}
$$
In particular, there exists $n_0\geq 1$ such that, for all $n\geq n_0$, $|S_n| \geq \frac{\alpha}{2}n$. Hereafter, we consider $n\geq n_0$. Now, since $a_n \geq 0$ for all $n$, we have
$$
\sum_{k=1}^n a_k \geq \sum_{k\in S_n} a_k\geq \sum_{k\in S_n} \frac{1}{k}\geq \frac{\alpha}{2}n\cdot \frac{1}{n} = \frac{\alpha}{2}\,.
$$
Of course, this is not enough to reach a conclusion. Let's try again: fix $n\geq n_0$, and for $t\in\mathbb{N}$ (recalling that  $S_n \subseteq S_{2n} \subseteq \dots\subseteq S_{tn}$), 
$$\begin{align}
\sum_{k=1}^{tn} a_k &\geq \sum_{k\in S_n} a_k + \sum_{k\in S_{2n}\setminus S_n} a_k+
 \dots
+ \sum_{k\in S_{tn}\setminus S_{(t-1)n}} a_k \\
&\geq \sum_{k\in S_n} \frac{1}{k} + \sum_{k\in S_{2n}\setminus S_n} \frac{1}{k}+ \dots + \sum_{k\in S_{tn}\setminus S_{(t-1)n}} \frac{1}{k} \\
&\geq \frac{|S_n|}{n} + \frac{|S_{2n}\setminus S_n|}{2n}+ \dots + \frac{|S_{tn}\setminus S_{(t-1)n}|}{tn} \\
&= \frac{|S_n|}{n} + \frac{|S_{2n}|-|S_n|}{2n}+ \dots + \frac{|S_{tn}|-  |S_{(t-1)n}|}{tn} \\
&= \frac{|S_n|}{2n}+ \frac{|S_{2n}|}{6n}+ \dots +  \frac{|S_{sn}|}{s(s+1)n}+ \dots +  \frac{|S_{(t-1)n}|}{t(t-1)n} + \frac{|S_{tn}|}{tn} \\
&\geq  \frac{\alpha}{2}\left(\frac{1}{2}+ \frac{1}{3}+ \dots +  \frac{1}{s+1}+ \dots +  \frac{1}{t} + 1\right) \\
&=  \frac{\alpha}{2} H_t\,.
\end{align}$$
the second-to last line since $S_{sn}$ has at least $\frac{\alpha}{2} sn$ elements.
Since the Harmonic series goes to infinity, we get
$$\begin{align}
\sum_{k=1}^{tn} a_k \geq  \frac{\alpha}{2} H_t \xrightarrow[t\to\infty]{} \infty\,,
\end{align}$$
contradicting convergence of the series.   $\square$
A: Another way to state your condition is
$$
\lim_{n\to\infty}\frac1n\sum_{j=1}^n\left[a_j\ge\frac1j\right]=0\tag1
$$
where $[\dots]$ are Iverson Brackets.
Suppose that $(1)$ is false, then there is an $\epsilon\gt0$ so that for infinitely many $n$,
$$
s_n=\sum_{j=1}^n\left[a_j\ge\frac1j\right]\ge n\epsilon\tag2
$$
Find an $n_1$ that satisfies $(2)$ great enough that $n_1\epsilon\ge1$. Then find $n_k$ that satisfies $(2)$ and $n_k\epsilon\ge2s_{n_{k-1}}$.
Since $s_{n_{k-1}}\le\frac12n_k\epsilon\le\frac12s_{n_k}$, we have $s_{n_k}-s_{n_{k-1}}\ge\frac12s_{n_k}$. Furthermore, 
$$
\sum_{n_{k-1}\lt j\le n_k}a_j\ge\frac{\frac12s_{n_k}}{n_k}\ge\frac12\epsilon\tag3
$$
Since the series is the sum over an infinite number of intervals like the one in $(3)$, the series diverges.
Assuming that $(1)$ is false leads to the conclusion that the series diverges. Thus, if the series converges, $(1)$ is true.
A: Assume $a_n\geq 0$ and $\sum_{n=1}^\infty a_n<\infty$ be a convergent series. 
Define $$b_n=|\{k: \frac{1}{n}\leq a_k<\frac{1}{n-1}\}|$$. 
We have that $$\sum_{n=1}^\infty\frac{ b_n}{n}\leq \sum_{n=1}^\infty a_n<\infty$$
We have $$\frac{|\{k\leq n:a_k\geq 1/k\}|}{n}\leq\frac{|\{k\leq n:a_k\geq 1/n\}|}{n} =\frac{1}{n}\sum_{k=1}^n {b_k}.$$
To prove what we want is enough to prove the following:
For $b_n\geq 0$.$$\hbox{If }\sum_{n=1}^\infty \frac{b_n}{n}=L<\infty\hbox{ then }\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n {b_k}=0$$
Proof: Let $A(x)=\sum_{n\leq x} \frac{b_n}{n}$. We have $\lim_{x\to\infty} A(x)=L$. Note 
$$\frac{1}{n}\sum_{k=1}^n {b_k}=\frac{1}{n}\int_{0}^{n^+}x\operatorname{d}(A(x))$$
By integration by parts(summation by parts) we have
$$\frac{1}{n}\sum_{k=1}^n {b_k}=A(n)-\frac{1}{n}\int_{0}^{n}A(x)\operatorname{d}x$$
Since $A(x)\to L$ as $x\to\infty$ we have that the above converges to $L-L=0$.
