Convergence of the series $\sum_ {n\geq1} \frac {(f(n) +P(n)) \pmod {Q(n)}} {D(n)}$

Solution 1:

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.