Convergence of Complex Series $\sum_{n=1}^{\infty}\frac{1}{q+n}$ Suppose I have a series given by $\sum_{n=1}^{\infty}\frac{1}{q+n},$ where $q \in \mathbb{C}.$ How does one go about determining the convergence/divergence of this series?
Idea: I thought to take the real part of the series. It would be roughly of the form $\frac{1}{n},$ which is a $p$-series with $p=1$, which is divergent. As a complex series converges iff its real and imaginary parts do, then I would conclude such a series to be divergent.
 A: You have
$$
\frac1{a+ib+n}=\frac{a+n-ib}{(a+n)^2+b^2}=\frac{a+n}{(a+n)^2+b^2}+i\,\frac{b}{(a+n)^2+b^2}.
$$
Then
$$
\sum_n\frac1{a+ib+n}=\sum_n\frac{a}{(a+n)^2+b^2}+i\,\sum_n\frac{b}{(a+n)^2+b^2}+\sum_n\frac{n}{(a+n)^2+b^2}.
$$
For any $q=a+ib$ the first two series on the right converge, while the last one diverges. Hence the original series always diverges.
A: $\frac1{q+n}-\frac1n=-\frac q{n(q+n)}$ and $\frac{|q|}{n|q+n|}\sim\frac1{n^2}$ hence $\sum_{n=1}^{\infty}\frac{1}{q+n}$ is divergent, as sum of a divergent (harmonic) series and a (absolutely) convergent one.
A: If $\sum_n a_n$ does not converge and $\sum_n b_n$ converges then $\sum_n (a_n+b_n)$ does not converge. Let $a_n=\frac {1}{n}.$ For $n>2|q|$ let $b_n=\frac {-q}{n(n+q)}.$ For $n>2|q|$ we have $|b_n|\le \frac {2|q|}{n^2}.$
A: We observe that $\frac {1}{q+n}\sim \frac 1n$, for $\frac {\frac {1}{q+n}}{\frac 1n}=\frac {n}{q+n}=\frac {1}{\frac qn +1}\to 1$. By the asymptotic equivalence with the harmonic series we conclude the divergence of the given series.
