Convergence of the series $\sum_ {n\geq1} \frac {(f(n) +P(n)) \pmod {Q(n)}} {D(n)}$ 
Under what conditions does the series $$\sum_ {n = 1}^\infty \frac {(f(n) +P(n)) \pmod{ Q(n)}} {D(n)}$$  converge?

Here $\text{P}$, $\text{Q}$ and $\text{D}$ are polynomials, with $\text{deg}(Q)= \text{deg}(D)-1$.   
Also $f(n)= \displaystyle \sum_{r=2}^{k} a_{r} r^n$ where $ \{a_{r} \}_{r=2}^k$ are real numbers and k is a natural number with k>2. k and the degree of P must be larger than the degree of Q. 
Here and here particular cases of this question have been answered by the user @SangchulLee. 
I'd like to know in which cases this series converges. (I assume they are rare but Sangchul Lee has already found some). I'm getting somewhere with this so don't flag me yet!
 A: As noted in the comments, heuristically the series will diverge if $f(n) \neq 0$. A literature search through equidistribution theorems would be the first thing to do if you wanted to prove it.
So, suppose $f(n) = 0$. Sangchul's classification for the $D(n) = n^2$ case generalizes to the present case without difficulty. The tweaks:

*

*$B$ is now $B(x)$, with $\deg B(x) < \deg Q(x) < \deg D(x)$.

*First case: $\{B(n)/Q(n)\}$ still becomes $B(n)/Q(n)$ for $n$ large and $\sum_{i=1}^\infty B(n)/D(n)$ still converges since $\deg D(x) - \deg B(x) \geq 2$.

*Second case: the lower bound becomes $\epsilon \sum_{k=1}^\infty \frac{Q(ak+b)}{D(ak+b)} + O(1)$. Since $\deg D(x) - \deg Q(x) = 1$, this still diverges by comparison with the harmonic series.

*Third case: the sum becomes $\sum_{n=1}^\infty \frac{Q(n)}{D(n)} \left\{\frac{P(n)}{Q(n)}\right\}$ where the term in braces remains equidistributed mod $1$. This again diverges by comparison with the harmonic series.

*Conclusion: the series $\sum_{n=1}^\infty \frac{P(n) \pmod{Q(n)}}{D(n)}$ with $\deg Q(x) = \deg D(x) - 1$ converges if and only if $P$ is of the form
$$P(x) = Q(x) \sum_{k=0}^n c_k \binom{x}{k} + B(x)$$
where $B(x)$ is a real polynomial with $\deg B(x) < \deg Q(x)$.

Hence there's no more interesting cases than the ones already found, so long as you believe the exponential heuristic.
