I started exploring the following question:
If $\ (a_n)_{n \in \mathbb{N}},\ $ is a sequence of natural numbers, and $\ \displaystyle\sum_n \frac{1}{a_n}\ $ diverges, then $\ \displaystyle\sum_n \frac{1}{a_n + n}\ $ diverges.
But then I found the following counter-example:
$$ a_n= \begin{cases} 1&\text{if}\, n=k^2,\ k\in\mathbb{N}\\ 2^{n}&\text{otherwise} \end{cases} $$
And there are similar counter-examples if we replace "$+n$" in the question with "$+\sqrt{n}$", or any other increasing function of $n$. I think the fact that $a_n$ is allowed to be non-increasing allows for this misbehaviour. Hence the natural follow-on question; is this true or false, and how to prove it:
If $\ (a_n)_{n \in \mathbb{N}},\ $ is a non-decreasing sequence of natural numbers, and $\ \displaystyle\sum_n \frac{1}{a_n}\ $ diverges, then $\ \displaystyle\sum_n \frac{1}{a_n + n}\ $ diverges.
I think, of course, we can replace "sequence of natural numbers" in these questions with "sequence of positive real numbers", and it's essentially the same question. So this distinction doesn't affect the question.
And of course if $\ a_n\ $ were required to be strictly increasing, which it is not, then the answer is affirmative, because then we would have, $\ \displaystyle\sum_n \frac{1}{a_n + n} \geq \displaystyle\sum_n \frac{1}{a_n + a_n} = \frac{1}{2}\displaystyle\sum_n \frac{1}{a_n},\ $ which diverges by supposition.