Recall that the harmonic series $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{n}}$ diverges.
This is because we may bound the partial sums below, like so:
$$1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots >$$ $$\frac{1}{2}+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\cdots=$$ $$\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots\to\infty$$
We may replace $n$ by $an+b$, where $a,b\in\mathbb{R}$, $a$ and $b$ not both $0$, to retain a divergent sum, which we will prove below.
Proposition. Let $a,b\in\mathbb{R}$ with $a$ and $b$ not both 0. Then, $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{an+b}}$ diverges.
Proof.
When $a=0$, the terms in the series are constant and non-zero, and so the sum diverges.
In the case $a<0$, multiplying by -1 gives a positive coefficient in front of $n$, so we may assume $a>0$.
For $b=0$, note that $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{an}=\frac{1}{a}\sum_{n=1}^{\infty}\frac{1}{n}}$ diverges.
When $b<0$, we have that $\frac{1}{an+b}>\frac{1}{an}>0$, and so $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{an+b}}$ diverges by the comparison test.
Finally, we consider the case $b>0$. Choose a positive integer $k>0$ such that $ak \geq b$. Then,
$$S_{N}=\sum_{n=1}^{N}\frac{1}{an+b}=\sum_{n=k+1}^{N+k}\frac{1}{a\left(n-k\right)+b}=$$ $$\sum_{n=k+1}^{N+k}\frac{1}{an + b – ka} = \sum_{n=k+1}^{N+k}\frac{1}{an + c},$$
where $c=b-ka\leq0$, which is covered in the cases above.
Therefore, $\displaystyle{\sum_{n=1}^{\infty}}\frac{1}{an+b}$ diverges for all $a,b\in\mathbb{R}$ with $a$ and $b$ not both 0.
QED
An interesting consequence of this is an example of an “almost alternating” series with $a_{n}\to0$ which diverges.
Recall that a series $\sum a_{n}$ is an alternating series if the sign changes every other term and converges (conditionally) if $a_{n}\to0$. In fact, as long as we eventually reach an alternating series after only finitely many terms in the sum, then it converges. Let’s test to see if alternating every other term is really so important.
Consider the “almost alternating” harmonic series, given by $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$. We may see that this sum diverges by grouping terms like so:
$$1+\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{1}{4}+\left(\frac{1}{5}-\frac{1}{6}\right)+\frac{1}{7}+\left(\frac{1}{8}-\frac{1}{9}\right)\cdots\gt$$ $$1+\left(\frac{1}{2}-\frac{1}{2}\right)+\frac{1}{4}+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{7}+\left(\frac{1}{8}-\frac{1}{8}\right)\cdots=$$ $$1+\frac{1}{4}+\frac{1}{7}+\cdots=\sum_{n=1}^{\infty}\frac{1}{3n-2},$$
which diverges.