Bounding the "slowest diverging" or "slowest converging" sum of reciprocals Let it be the infinite set of positive integers $S=\{{a_1,a_2,...}\}$ such that $\sum_{i=1}^{n}a_i=\lfloor\frac{n^2\sqrt(n)}{\ln\left({n}\right)}\rfloor$. Does the sum $\sum_{i=1}^{\infty}\frac{1}{a_i}$ converge or diverge? If it converges, to which limit? And if it diverges, at which rate?
I find interesting this series in particular because if we consider the infinite set of positive integers $S=\{{b_1,b_2,...}\}$ such that $\sum_{i=1}^{n}b_i=\lfloor{n^2\sqrt(n)}\rfloor$, it can be proved that the sum $\sum_{i=1}^{\infty}\frac{1}{b_i}$ converges, as $b_n\sim\frac{n^2+7n+2}{2}$; and other hand, if we consider the infinite set of positive integers $S=\{{c_1,c_2,...}\}$ such that $\sum_{i=1}^{n}c_i=\lfloor{n^2\ln(n)}\rfloor$, it can be proved that the sum $\sum_{i=1}^{\infty}\frac{1}{c_i}$ diverges, as $c_n\sim n\ln(n)$.
Additionally, I am interested in it because of this other post I published (Question on convergence / divergence of sums of reciprocals of positive integers); I am trying to bound as sharply as possible the proposed function $F(n)$. So if you have any idea of how could it be done, is more than welcomed.
Thanks in advance!
 A: If $S_n = \dfrac{n^{2.5}}{\ln(n)}$, we have $$a_n \sim S_n -S_{n-1} \sim \frac{5 \ln(n)-2}{2 \ln(n)^2} n^{3/2}$$
In particular, if $1 < p < 3/2$, $a_n > n^{p}$ for sufficiently large $n$, so $\sum_n 1/a_n$ converges.
EDIT: The asymptotics on $S_n - S_{n-1}$ arise this way.
$$\eqalign{S_{n-1} &= \dfrac{(n-1)^{5/2}}{\ln(n-1)} = \dfrac{n^{5/2}(1-1/n)^{5/2}}{\ln(n) + \ln(1-1/n)} \sim \dfrac{n^{5/2} - (5/2) n^{3/2})}{\ln(n) - 1/n}\cr & \sim \left(n^{5/2} - \frac{5}{2} n^{3/2}\right) \left( \frac{1}{\ln(n)} +
\frac{1}{n \ln(n)^2}\right)\cr &\sim \frac{n^{5/2}}{\ln(n)} - \frac{5}{2} \frac{n^{3/2}}{\ln(n)} + \frac{n^{3/2}}{\ln(n)^2} }$$
A: Some trivial observations
Let $a_n>0$, set $S_n=\sum_{k=1}^n a_k$ and let $F_n>0$.
Assume that:

*

*(A) $a_n$ is increasing ($a_n\le a_{n+1}$).

*(B) there exist an $L>0$ and $n_L$ such that, if $n>n_L$
$$
0<L<\frac{S_n}{F_n}
$$

*(C) $\sum_{n>n_L} \frac{n}{F_n}<\infty$
Then
$$
\sum_{n} \frac{1}{a_n}<\infty
$$
"Proof":
If $n>n_L$:
$$
0<L\stackrel{(B)}{<}\frac{S_n}{F_n}\stackrel{(A)}{\le} \frac{n a_n}{F_n} \implies \\
\frac{1}{a_n}<\frac{2}{L F_n}
$$
By comparison (C) implies the convergence.
Consequently, if $a_n$ is an increasing integer sequence and $\frac{S_n}{F_n^{(k)}}\to 1$ where $F_n^{(k)}$ is
one of the sequences below
$$
F_n^{(0)}=n\cdot n^{p}\\
F_n^{(1)}=n\cdot n\log(n)^p\\
F_n^{(2)}=n\cdot n\log(n)\log(\log(n))^p\\
F_n^{(3)}=n\cdot n\log(n)\log(\log(n))\log(\log(\log(n)))^p\\
...
$$
where p>1, then $\sum_n \frac{1}{a_n}<\infty$.
It means that we have a sequence of possible bounds with $F^{(k)}_n>F^{(k+1)}_n$, moreover
$\frac{F^{(k)}_n}{F^{(k+1)}_n} \stackrel{n\to \infty}{\to} \infty$, which suggests that there is no optimal bound. (But I do not see a general way to assess this claim.)
