Let it be the infinite set of positive integers $S=\{{a_1,a_2,…}\}$ such that $\sum_{i=1}^{n}a_i=\lfloor{n^2\ln\left({n}\right)}\rfloor$. Does the sum $\sum_{i=1}^{\infty}\frac{1}{a_i}$ converge or diverge? And if it converges, to which limit?
I find interesting this series in particular because the sum $\sum_{i=1}^{\infty}\frac{1}{p_i}$ of reciprocals of prime numbers diverges at a rate of aproximately $\ln\ln\left({n}\right)$, and $\sum_{i=1}^{n}p_i\approx\frac{1}{2}n^2\ln\left({n}\right)$. Therefore, if the series proposed diverges, it is expected to do so at a very slow divergence rate.
Thanks in advance!
Best Answer
Hint: $a_n=[n^{2} \ln (n)]-[(n-1)^{2}\ln (n-1)] \leq n^{2} \ln (n)+1-(n-1)^{2}\ln (n-1)$. Now $n^{2} \ln (n)-n^{2} \ln (n-1)=n^{2}\ln (1+\frac 1 {n-1})\leq \frac {n^{2}} {n-1}$. Now you can easily see that $a_n =O(n \ln (n))$. Since $\sum \frac 1 {n \ln (n)}$ is divergent it follows that $\sum \frac 1 {a_n}=\infty$.