2
$\begingroup$

It is known that if a set of natural numbers has positive asymptotic density then the sum of the reciprocals of those elements diverge. Let $\{a_n\}$ be an increasing sequence of natural numbers where the density of the set $\{a_n | n\in \mathbb N\}$ does not exist. Can the series $\sum_{i=0}^n \frac{1}{a_i}$ converge? All the examples that I have of sets where the density does not exist do not converge.

$\endgroup$

marked as duplicate by Aryabhata, N. F. Taussig, user147263, Surb, Community Mar 26 '15 at 14:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

No, it cannot. If the density of the sequence $(a_j)$ does not exist (or exists and is positive), then there must be some $M\in\mathbb{Z}_{>0}$ such that $$\limsup_{n\to\infty} \phi(n)>\frac 2 M, \qquad {\rm where\ \ } \phi(n):=\frac{\#(\{1,\ldots,n\}\cap \{a_j\})}{n}.$$ Therefore there is some sequence $m_1$, $m_2$, $\dots$ such that $m_k\to\infty$ and $\phi(m_k)> 2/M$ for each $k$. Pass to a subsequence so that $m_{k+1}>M m_k$ for each $k$. Then $\{m_k+1,m_k+2,\ldots,m_{k+1}\}\cap \{a_j\}$ has cardinality at least $$ \frac{2}{M} m_{k+1} - m_k > \frac{1}{M} m_{k+1} $$ so the sum of the reciprocals of the elements of $\{m_k+1,m_k+2,\ldots,m_{k+1}\}\cap \{a_j\}$ is at least $$ \frac{1}{m_{k+1}-\lfloor m_{k+1}/M \rfloor} + \cdots + \frac{1}{m_{k+1}-1} + \frac{1}{m_{k+1}}=\log \frac{M}{M-1} + o(1). \ \ \ (*) $$ Summing $(*)$ over $k$ then shows that $\sum_j a_j^{-1}$ diverges.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.