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!
Best Answer
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. $$\eqalign{S_{n-1} &= \dfrac{(n-1)^{5/2}}{\ln(n-1)} = \dfrac{n^{5/2}(1-1/n)^{5/2}}{\ln(n) + \ln(1-1/n)} \sim \dfrac{n^{5/2} - (5/2) n^{3/2})}{\ln(n) - 1/n}\cr & \sim \left(n^{5/2} - \frac{5}{2} n^{3/2}\right) \left( \frac{1}{\ln(n)} + \frac{1}{n \ln(n)^2}\right)\cr &\sim \frac{n^{5/2}}{\ln(n)} - \frac{5}{2} \frac{n^{3/2}}{\ln(n)} + \frac{n^{3/2}}{\ln(n)^2} }$$