I am asked to show that a complex series converges or diverges. (Based on the question I am asked, I assume it converges, but I am not too sure.) I have broken it down into four components:
- The basic initial terms
- A negative real series
- A positive real series
- An imaginary series
I found on StackExchange that you can take doubly alternating series (++--++-- etc.) and break them up into a negative and a positive series, which I did with (2) and (3). I am left with two series of the form
$$S_1 = (1/13+1/19+1/23+1/29+⋯)$$
and the series
$$S_2 = (1/7+1/17+⋯)$$
which I need to show converges. Here the series $S_1$ consists of the sum of the inverses of all prime numbers $\equiv 3 mod(5)$ and $\equiv 4 mod(5)$. Here the series $S_2$ conists of the sum of the inverses of all (odd) prime numbers ending in the digit 7 (that is, $\equiv 2 mod(5)$).
Neither $S_1$ nor $S_2$ converges individually, but a suitable version of $S_1-S_2$ converges.
Define \begin{align*} f_+(n) &= \begin{cases} 1, & \text{if $n\equiv1$ (mod $5$) or $n\equiv2$ (mod $5$),} \\ 0, & \text{otherwise;} \end{cases} \\ f_-(n) &= \begin{cases} 1, & \text{if $n\equiv3$ (mod $5$) or $n\equiv4$ (mod $5$),} \\ 0, & \text{otherwise;} \end{cases} \\ f(n) &= f_+(n) - f_-(n). \end{align*} Then $\displaystyle S_1 = \sum_p \frac{f_+(p)}p$ and $\displaystyle S_2 = \sum_p \frac{f_-(p)}p$, where the sums are taken over all primes $p\ne5$ (including $2$, $3$, and $11$, even though they're not present in the OP's examples). The fact that neither one of these sums converges follows from the fact that whenever $\gcd(a,q)=1$, the sum $\displaystyle \sum_{p\equiv a\text{ (mod }q)} \frac1p$ diverges (as Dirichlet proved in 1837); indeed an asymptotic formula is known for its rate of divergence.
However, suppose we defined $S = \displaystyle\sum_p \frac{f(p)}p$, which is a suitable interpretation of $S_1-S_2$. If we define $\chi$ to be a complex Dirichlet character modulo $5$, namely $$ \chi(n) = \begin{cases} 0, & \text{if $n\equiv0$ (mod $5$),} \\ 1, & \text{if $n\equiv1$ (mod $5$),} \\ i, & \text{if $n\equiv2$ (mod $5$),} \\ -i, & \text{if $n\equiv3$ (mod $5$),} \\ -1, & \text{if $n\equiv4$ (mod $5$),} \end{cases} $$ then we can check that $f(n) = \frac{1-i}2\chi(n) + \frac{1+i}2\bar\chi(n)$. We can thus write $$ \sum_p \frac{f(p)}p = \frac{1-i}2 \sum_p \frac{\chi(p)}p + \frac{1+i}2 \sum_p \frac{\bar\chi(p)}p; $$ and it is known that both series on the right-hand side converge (essentially because $L(1,\chi)$ converges and is nonzero).