Convergence of series

Let $\mathrm{gpf}$ denote the greatest prime factor of $n$ . For example $\mathrm{gpf}(17) = 17$ , $\mathrm{gpf} (18) =3$ . Define $\mathrm{gpf}(1)=1$ . Examine if the sum

$\mathcal{S} = \sum_{n=1}^{\infty} \frac{1}{n \; \mathrm{gpf}(n)}$

converges.

Solution

Let $p_k$ be the $k$ -th prime number.

$\sum_{n\ge1}\frac{1}{n\operatorname{gpf}(n)}=\sum_{k \geq 1}\frac{1}{p_k}\sum_{\operatorname{gpf}(n)=p_k} \frac{1}{n}$

If $\operatorname{gpf}(n)=p_k$ , then $n=p_1^{i_1}\dots p_{k-1}^{i_{k-1}}\,p_k^{i_k}$ with $i_j \geq 0$ , $1 \leq j<k$ and $i_k \geq 1$ . It folows that

$\begin{align*} \sum_{\operatorname{gpf}(n)=p_k}\frac{1}{n}&=\left(\sum_{i_1 \geq 0}p_1^{-i_1} \right)\cdots \left(\sum_{i_{k-1}\geq 0} p_{k-1}^{-i_{k-1}} \right) \left(\sum_{i_k\ge1}p_k^{-i_k}\right)\\ &=\frac{1}{p_k}\,\prod_{i=1}^k \left(1-\frac{1}{p_i} \right)^{-1} \end{align*}$

From Merten’s theorem

$\prod_{i=1}^k \left(1-\frac{1}{p_i} \right)^{-1}\sim e^\gamma\,\log p_k$

and the original series has the same character as

$\sum_{k=1}^{\infty} \frac{\log p_k}{p_k^2}$

which is convergent.

Convergence of series

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