Home » Uncategorized » Convergence of series

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

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



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.

Read more

Leave a comment

Donate to Tolaso Network