1

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!

Juan Moreno
  • 1,110
  • What is $a_1$? For $n=1$ it should satisfy $$a_1=\sum_{i=1}^na_i=\lfloor\frac{1^2\cdot\sqrt{1}}{\ln1}\rfloor,$$ which is not defined. – Servaes Aug 27 '20 at 15:40
  • And in general if $\sum_{i=1}^nc_i\sim n^k$ for some $k>2$, then $\sum_{i\geq1}\tfrac{1}{c_i}$ converges. – Servaes Aug 27 '20 at 15:44

2 Answers2

1

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

Robert Israel
  • 448,999
  • thanks for your answer! Could you elaborate a bit in the approximation you get, and in the reasoning you make to prove convergence? And, digging further, which would you say it would be a sharp bound of the function $F(n$ proposed? – Juan Moreno Aug 27 '20 at 08:32
1

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.)

Czylabson Asa
  • 328