Here is an interesting candidate for the slowest diverging family of sequences. It is well known that every $a_n$ in the family

$$n\log^{2}(n)$$

$$n\log(n)\log^2(\log(n))$$

$$n\log(n)\log(\log(n))\log^2(\log(\log(n)))$$

$$...$$

has the property that they converge when summed reciprocally, so to have any chance of being the fastest it would need to be able to beat all of these. In this discussion, we will assume that all logarithms are taken base $\delta$ (for some fixed delta) which we will come back to choosing later. We will also use the notation

$$\log^{+}(x)=\max(1,\log(x))$$

so that $\log^+(\log^+(\log^+(...(\log^+(x)))))$ is always defined and real. We also let $\log^{(m)}(x)$ denote the $m-$fold application of the $\log^{+}$ function on $x$. We now define

$$\mathscr{L}_k(x):=n\left(\prod_{n\leq k}\log^{(n)}(x)\right)\log^{(k)}(x)$$

so that $\mathscr{L}_0(x)=n$, $\mathscr{L}_1(x)=n\log^2(n)$, etc... We now defined the very natural "closure of these" functions

$$\mathscr{L}_{\infty}(x):=\lim_{k\to\infty}\mathscr{L}_k(x)$$

note that we dropped the extra $\log^{(k)}(x)$ factor when taking the limit, since it will always approach $1$, so

$$\lim_{x\to\infty}\frac{\mathscr{L}_{\infty}(x)}{\mathscr{L}_{k}(x)}=0$$

for any choice $k$. The proof of the above statement is more complicated than the handwaving in the present answer, but it is not very hard at all. Before showing that

$$\sum_{n}\frac{1}{\mathscr{L}_{\infty}(n)}$$

converges we need some preliminaries. We begin by noticing that for $x>1$

\begin{align*}
\mathscr{L}_{\infty}\left(\delta^x\right)&=\delta^x\left(\prod_{n=1}^{\infty}\log^{(n)}\left(\delta^x\right)\right)\\
&=\delta^x x\left(\prod_{n=1}^{\infty}\log^{(n)}\left(x\right)\right)\\
&=\delta^x\mathscr{L}_{\infty}(x)\tag{1}
\end{align*}

where we used that for $m>1$ $\log^{(m)}\left(\delta^x\right)=\log^{(m-1)}\left(x\right)$ uniformly, and $\log^{+}\left(\delta^x\right)=x$ since $x>1$. We now move on to proving the following theorem:

The series
\begin{equation}
\sum_{n}\frac{1}{\mathscr{L}_{\infty}\left(n\right)}\tag{2}
\end{equation}
converges $\mathbf{if\,\,and\,\,only\,\,if}$ $\delta<e$.

To do this, we use the integral test to see that the series in (2) converges if and only if the integral

$$I(x)=\int_{\delta}^{x}\frac{1}{\mathscr{L}_{\infty}\left(t\right)}dt$$

converges as $x\to\infty$. Using basic calculus tricks we see that for $x>\delta$

\begin{align*}
I\left(x\right)&=\int_{\delta}^{x}\frac{1}{\mathscr{L}_{\infty}\left(t\right)}dt\\
&=\int_{1}^{\log(x)}\frac{\ln(\delta)\delta^t}{\mathscr{L}_{\infty}\left(\delta^{t}\right)}dt\\
&=\int_{1}^{\log(x)}\frac{\ln(\delta)}{\mathscr{L}_{\infty}\left(t\right)}dt\\
&=\int_{1}^{\delta}\frac{\ln(\delta)}{\mathscr{L}_{\infty}\left(t\right)}dt+\int_{\delta}^{\log(x)}\frac{\ln(\delta)}{\mathscr{L}_{\infty}\left(t\right)}dt\\
\end{align*}

for $1<x<\delta$, $\log(x)<1$ and so $\mathscr{L}_{\infty}\left(t\right)=t$, meaning that

$$I(x)=\ln^2(\delta)+\ln(\delta)I(\log(x))$$

if $\log^{(m)}(x)\geq1$ but $\log^{(m+1)}(x)<1$, then we can apply the above formula $m$ times to get that

\begin{align*}
I(x)&=\sum_{j=1}^m\ln^{j+1}(\delta)+\ln^m(\delta)I\left(\log^{(m)}(x)\right)\\
&=\sum_{j=1}^m\ln^{j+1}(\delta)+O\left(\ln^m(\delta)\right)
\end{align*}

thus, as $x\to\infty$, $I(x)$ will converge if and only if the series

$$\sum_{j=1}^m\ln^{j+1}(\delta)$$

converges which will take place only when $\ln(\delta)<1\implies \delta<e$. This completes our proof.

One interesting thing to notice is that for any of the finite $\mathscr{L}_{k}\left(x\right)$ the base does not matter since they will all be off by at most a constant factor. A reason this family of functions $\mathscr{L}_{\infty}\left(x\right)$ is such a strong candidate is that we have endpoint failure at $\delta\geq e$, which shows that we have a very small space to work with when improving.