Help with proof that $\sum_{n \in \Bbb{N}} \frac{1}{an + b}$ also diverges? We know that $\sum_{n \in \Bbb{N}} \frac{1}{n}$ diverges.  So it seems likely that $\sum_{n \in \Bbb{N}} \frac{1}{a n  + b}$ will for any real $a, b$.  I'm having trouble proving it just for the $\frac{1}{n + c}$ form special case.  Any hints?
 A: Let $a,b \in \mathbb{R}$ and let $a \neq 0$. Since 
$$
\frac{1}{an+b} \sim \frac{1}{an}
$$
as $n$ grows,
that is to say,
$$
\frac{n}{an+b} \to \frac{1}{a}
$$
as $n$ grows,
so by the limit comparison test the series $\sum 1/(an+b)$ diverges. 
A: Hint(s):
$$\sum_{n \in \mathbb N} \frac{1}{n+c}=\sum_{n \in \mathbb N} \frac{1}{n}- \sum_{k=1}^c \frac {1}{k}$$
And $\frac{1}{an+b}$ can be rewritten as $\frac{1}{a} \cdot \frac{1}{n+\frac{b}{a}}$.
Now let $\frac{b}{a}:=c$.
A: First, since
$\sum_{n=1}^M \frac1{n}
\to \infty
$
as $M \to \infty
$,
then,
for any $m > 0$,
$\sum_{n=m}^M \frac1{n}
\to \infty
$
as $M \to \infty
$.
Do you see why?
Now,
let's look at the partial sums
of the problem's series.
$\sum_{n=1}^M \frac1{an+b}
=a\sum_{n=1}^M \frac1{n+b/a}
$.
Let $k$ be an integer
such that $k \ge b/a$.
Then
$\frac1{n+b/a}
\ge \frac1{n+k}
$,
so
$\sum_{n=1}^M \frac1{n+b/a}
\ge \sum_{n=1}^M \frac1{n+k}
= \sum_{n=k+1}^{M+k} \frac1{n}
$.
By the comment above,
$\sum_{n=k+1}^{M+k} \frac1{n}
\to \infty
$
as
$M \to \infty
$.
Therefore
$\sum_{n=1}^M \frac1{an+b}
\to \infty
$
as $M \to \infty
$.
A: It very simple to proof that if $(*)\frac{|u_n|}{|v_n|}\to L\neq 0,\infty$ so 
$$\sum_n |u_n|<\infty \qquad \iff\qquad \sum_n |v_n|<\infty$$
in fact $(*)$ implies that : for all $L>\epsilon>0$ it exist $N_0\in\mathbb{N}$ such that $\forall n\geq N_0$  $\left| \frac{|u_n|}{|v_n|}- L\right|<\epsilon$ and this implies that for  $N_0\in\mathbb{N}$ 
$$
(L-\epsilon)|v_n|<|u_n|<(L+\epsilon)|v_n|
$$
So 
$$
(L-\epsilon)\sum_{n\geq N_0}|v_n|\leq \sum_{n\geq  N_0} |u_n|\leq (L+\epsilon)\sum_{n\geq N_0}|v_n|
$$
and this proof our result.
Now we applies this result to your case :
where $v_n=n$ and $u_n=an+b$.
So $\frac{|u_n|}{|v_n|}\to a$
if $a>0$ so $\exists N \in \mathbb{N}$ such that $v_n\geq0$ for all $n\geq N$ and then $\sum_n v_n =\infty$.
If $a<0$ so $\exists N' \in \mathbb{N}$ such that $-v_n\geq0$ for all $n\geq N'$ and then $-\sum_n v_n =\infty$.
In booth cases $\sum_n\frac{1}{an+b}$ diverge for all $a,b\in\mathbb{R}$
A: Here's a quick one. $an + b \leq an + bn = (a + b) n$, so
\begin{align*}
\sum_{n = 1}^{\infty} \frac{1}{an + b} & \geq \sum_{n = 1}^{\infty} \frac{1}{an + bn} \\
& = \frac{1}{a + b} \sum_{n = 1}^{\infty} \frac{1}{n} \\
& = \frac{+ \infty}{a + b} \\
& = + \infty .
\end{align*}
