If $\sum a_n$ converges , then $a_n<1/n$ a.e?

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. $\square$

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.

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 $L-L=0$.