Is $\sum\limits_{n=1}^{\infty}p_n$ convergent? The sequence $\{p_n\}$ is defined as follows:
$$p_1=\frac{b}{a},$$
$$p_n=\frac{b-c\sum\limits_{i=1}^{n-1}p_i}{na}, \qquad (n\geq 2).$$
where a, b and c are parameters satisfying $a\gt b\gt 1\gt c$.
Is $\sum\limits_{n=1}^{\infty}p_n$ convergent or divergent? Does it depend on the values of a, b and c?

Edit
Alright, to answer the "close" requests, let me provide the context (at the risk of introducing mathematically unnecessary and irrelevant details).
This problem arises when I was considering some one dimensional pursuit-evade problem. Specifically, the pursuer and evader both have speed 1, and the pursuer lags $b$ behind in the beginning. However, at every distance $a$ they travel, the pursuer gets a boost that propels him a certain distance towards the evader. That distance is proportional to the current distance between them, and inversely proportional to the total distance he has traveled. So for example, at the $a$ milestone, the pursuer gets a boost of $c\frac{b}{a}$; at the $2a$ milestone, he gets a boost of $c\frac{b-\frac{bc}{a}}{2a}$, etc. I was considering small $c$ and large $a$, wondering if the pursuer can ever catch up, or if the evader can keep him at bay for large $b$ (which means the summation of $p_n$ converges).
I tried to simplify $p_n$, but has made no progress so far.
 A: It follows easily from the recurrence that
$$
(n + 1)p_{n + 1}  = \left( {n - \frac{c}{a}} \right)p_n ,
$$
i.e.,
$$
\frac{{p_{n + 1} }}{{p_n }} = \frac{{n - \frac{c}{a}}}{{n + 1}} = 1 - \frac{{a + c}}{a}\frac{1}{n} + \frac{{a + c}}{a}\frac{1}{{n^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^3 }}} \right).
$$
By the Gauss criterion, the series $\sum p_n$ converges if and only if $\frac{{a + c}}{a} > 1$, i.e., if $c>0$.
A: We will prove that $\sum\limits_{n=1}^\infty p_n$ is convergent. It is easy to derive the rational dependence$$np_n=\frac ba-\frac ca\left(\sum_{i=1}^{n-2}p_i+p_{n-1}\right)=\left(n-1-\frac ca\right)p_{n-1}$$ from which $$p_n=p_1\prod_{k=2}^n\frac{k-1-c/a}{k}=\frac ba\frac1{n!}\prod_{k=2}^n\left(k-1-\frac ca\right).$$ Direct substitution gives $$\sum_{n=1}^\infty p_n=\frac ba\sum_{n=1}^\infty\left[\frac1n\prod_{k=2}^n\left(1-\frac c{(k-1)a}\right)\right]:=\frac ba\sum_{n=1}^\infty a_n$$ so it suffices to study the convergence of $\sum\limits_{n=1}^\infty a_n$. D'Alembert's ratio test proves inconclusive so we try Raabe's test in the De Morgan hierarchy. We have $$\rho_n=n\left(\frac{a_n}{a_{n+1}}-1\right)=n\left(\frac{n+1}{n-c/a}-1\right)\stackrel{n\to\infty}\to1+\frac ca>1$$ so convergence is determined. This conclusion does not depend on any of $a,b,c$ as long as $c>0$.
