A convergence test similar to Gauss' test. Consider a sequence of complex numbers $(a_n)$ and assume that we can write $$\frac{a_{n+1}}{a_n}=1+\frac{\lambda}n+\frac{b_n}{n^2}$$ where $b_n$ is bounded and $\Re(\lambda)<-1$. Can we show that $\sum_{n\geqslant 0}|a_n|<\infty$? If not, can the claim be fixed?
 A: Choose $n_0$ large enough so that $\left\lvert \frac{a_{n+1}}{a_n}-1\right\rvert < \frac{1}{4}$ (or anything small enough that we have $\lvert\log (1+u)-u\rvert \leqslant \lvert u\rvert^2$) for $n \geqslant n_0$. Then take logarithms.
$$\log \frac{a_{n+1}}{a_n} = \frac{\lambda}{n} + O\left(\frac{1}{n^2}\right),$$
so for a suitable branch of the logarithm, we have
$$ \log \frac{a_{n+1}}{a_{n_0}} = \lambda\log \frac{n}{n_0} + O(1),$$
and hence
$$a_{n+1} = a_{n_0}\cdot n^{\lambda}\cdot O(1),$$
which gives the majorant
$$\lvert a_{n+1}\rvert \leqslant Cn^{\Re\lambda}$$
for some $C < \infty$.
A: As
$$
\frac{a_{n+1}}{a_n}=1+\frac{\lambda}{n}+\frac{b_n}{n^2}\in\mathbb C,
$$
with $b_n$ bounded, and 
$$
\left(1+\frac{1}{n}\right)^{-\lambda}=\left(\frac{n+1}{n}\right)^{-\lambda}=1-\frac{\lambda}{n}+\frac{c_n}{n^2},
$$
$c_n$ also bounded, then
$$
\frac{(n+1)^{-\lambda}a_{n+1}}{n^{-\lambda}a_n}=1+\frac{d_n}{n^2},
$$
$d_n$ also bounded. Note that $f(z)=(1+z)^\alpha$ makes sense a holomorphic function for $\lvert z\rvert<1$, for every $\alpha\in\mathbb C$, and we amy assume that $f(0)=1$. 
Hence
$$
a_n=a_{n_0}\prod_{k=n_0}^{n-1}\frac{a_{k+1}}{a_{k}}=a_{n_0}e_n\,{n_0^{-\lambda}}
{n^{\lambda}}
$$
where
$$
e_n=\prod_{k=n_0}^n \left(1+\frac{d_k}{k^2}\right)
$$
and $e_n$ absolutely bounded from above and below. Altogether
$$
a_n=z_n\,n^{\lambda},
$$
where $z_n$ is absolutely bounded from both below and above, and hence, 
$$
k_1\,n^{\mathrm{Re}\,\lambda}\le \lvert a_n\rvert\le k_2\,n^{\mathrm{Re}\,\lambda},
\quad n\in\mathbb N,
$$
for some $0<k_1\le k_2$.
Hence $\{a_n\}$ is absolutely summable for Re$\,\lambda<-1$, and absolutely divergent if
Re$\,\lambda>-1$.
