I'm trying to prove the following statement (which may or may not be true):
Let $\{x_n\}$ denote a positive sequence of real numbers and suppose the sequence $\left\{\frac{x_n}{n}\right\}$ increases monotonically to infinity. Then there exists a non-negative sequence $\{c_n\}$ such that $$\sum_{n=1}^\infty x_nc_n = \infty \quad \text{ and } \quad \sum_{n=1}^\infty n c_n < \infty$$
Is the above statement true, or do I need more information about the rate of increase of $\left\{\frac{x_n}{n}\right\}$?
Thank you!
Tidying up the problem by calling $\frac{x_n}{n}=b_n, nc_n=a_n$, we have: if $b_n$ increases monotonically to infinity, find $a_n \ge 0, \Sigma{a_nb_n}=\infty, \Sigma{a_n} <\infty$ and that is very easy without any extra regularity conditions required on $a_n$ (like decreasing or such).
Wlog we can take $b_1 \ge 1$ and then we find an increasing sequence of integers $0=k_0 < k_1 <.... $ such that in $[2^{k_r}, 2^{k_{r+1}})$ there are $n_r \ge 1$ of the $b_n$'s.
(just take $k_1, 2^{k_1}> b_1$ and then $k_2>k_1$, s.t. $2^{k_2}>b_r \ge 2^{k_1}$ for some $r$ etc, so in other words we follow the $b$'s until they break $2^{k_r}$ and then take the next power of two bigger than the corresponding $b$ which can be high if the $b$'s spread out like factorials for example)
Then choosing $a_r=\frac{1}{n_r2^{k_r}}$ for the $b$'s in the $[2^{k_r}, 2^{k_{r+1}})$ interval, obviously the sum of the $ab$'s there is at least $1$ (and there infinitely many such, so the required divergence), while the sum of the $a$'s there is $\frac{1}{2^{k_r}}$ (so dominated by the geometric series of a half, hence convergence) and we are done.