$0 \leq u_{n} \leq \frac {c}{n} \sum\limits_{k=1}^{n-1} u_{k}$ $\implies$ $\sum\limits_{k \geq 0} u_{k} < + \infty$? In this question it is proved that: $0\leq u_n\leq \frac {1}{n^2}\sum_{k=1}^nu_k\implies $ $\sum u_k$ converges.
I would be interesting in the question that whether we can relax the assumption $$ 0 \leq u_{n} \leq \frac {1}{n^{2}} \sum\limits_{k=1}^{n} u_{k} $$ 
to
$$ 0 \leq u_{n} \leq \frac {c}{n} \sum\limits_{k=1}^{n-1} u_{k} $$ 
for some constant $c \in \left( 0 , 1 \right)$. Note that we also has one term less than the previous (the series only run till $u_{n-1}$).
We can not use the same approach as in the previous question since $\sum\limits_{k \geq 1} u_{k} = + \infty$.
 A: I am ignoring first few values of $n$ in the  hypothesis. $u_n=\log(1+\frac 1 {n+2})$ is a counterexample. Note that $\sum\limits_{k=1}^{n-1} u_k=\sum\limits_{k=1}^{n-1}[log(k+3)-log(k+2)]=\log(n+2)-\log\, 3$ from which you can see that the hypothesis is satisfied. 
A: With the convention that an empty sum evaluates to zero, we have for $n=1$
$$
 0 \le u_1 \le \frac c1 \sum_{k=1}^0 u_k = 0
$$
which implies $u_1= 0$. Then $u_n= 0$ for all $n\in \Bbb N$ follows via induction.
So the only sequence satisfying the hypothesis is the zero sequence, so that $\sum_{k \geq 0} u_{k}$ is (trivially) finite.

If the hypothesis is relaxed to
$$ \tag{*}
0 \leq u_{n} \leq \frac {c}{n} \sum_{k=1}^{n-1} u_{k} \text{ for sufficiently large $n$}
$$
then for any $c \in (0, 1)$ a counter-example is given by
$$ 
u_n = (n+1)^d - n^d
$$
where $d$ is chosen in $(0, c)$. From the mean-value theorem we can estimate the left-hand side as
$$
 u_n = (n+1)^d - n^d \le d n^{d-1}
$$
so that $(*)$ is satisfied if
$$
 d n^{d-1} \le \frac {c}{n} \sum_{k=1}^{n-1} u_{k} = \frac {c}{n} \bigl( n^d - 1\bigr) \\
\iff c \le (c-d)n^d
$$
and that is true for sufficiently large $n$.
A: Ignoring the condition for $n=1$, the assumption is incorrect for any $c > 0$.
To see this, define $u_1:=1$ and
$$u_n:=\frac{c}n\sum_{k=1}^{n-1}u_k, \text{ for } n\ge 2.$$
(REMARK: See below for a simplification of this proof).
This satisfies the condition of the problem, we have greedily chosen $u_n$ to be the maximal available value. 
It follows for $n \ge 2$ that $u_n \ge \frac{c}n u_1 = \frac{c}n$, as the remaining terms of the sum are non-negative.
Thus we get 
$$\sum_{k=1}^nu_k \ge \sum_{k=2}^nu_k \ge \sum_{k=2}^n\frac{c}k = c\sum_{k=2}^n\frac1k.$$
The sum on the right hand side is the known divergent harmonic series, and with $c > 0$ our sum is divergent by the comparison test.

SIMPLIFICATION: 
After $u_1:=1$, simpy continue with the explicit definition
$$u_n:=\frac{c}n, \text{ for } n\ge 2.$$
Since $u_n=\frac{c}nu_1$, the conditions is fullfilled for all $n\ge 2$ and divergence follows the same way as above.
