Does $-\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\frac{1}{19}+\frac{1}{23}+\dots$ converge?

Question

Let $d(m)$ be the number of positive divisors of $m$ [including $1$ and $m$]. Let $p_k$ be the $k^\text{th}$ prime number. Consider the series

$$\sum_{k=1}^{\infty }\frac{(-1)^{d(p_{k}-1)}}{p_{k}}=-\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\frac{1}{19}+\frac{1}{23}+\dots$$

Does the series converge? If yes, then what is its value of convergence?

Any help/hint will be appreciated. THANKS!

Answer 1

Since $d(n)$ is odd iff $n$ is a square, your sum is equal to the sum of reciprocals of primes, minus twice the sum of reciprocals of primes of the form $n^2+1$. But we have $$\sum_{p=n^2+1\text{ for some $n$}}\frac{1}{p}<\sum_{n=1}^\infty\frac{1}{n^2+1}<\infty,$$ so your sum diverges to infinity.