Do absolute convergence of $a_n$ implies convergence of $K_n=\frac{1}{\ln(n^2+1)}\sum_{k=1}^{+\infty}a_k\frac{3k^3-2k}{7-k^3}\sin k$? The problem asks us to decide whether the following statement is true:
Let $\{a_n\}_{n\geq1}$ be any absolutely convergent sequence. Does that imply that the sequence:
$$K_n=\frac{1}{\ln(n^2+1)}\sum_{k=1}^{n}a_k\frac{3k^3-2k}{7-k^3}\sin k$$
is convergent?
The term $\frac{1}{\ln(n^2+1)}$ goes to $0$ so if we showed that the series is convergent we would finish this problem. The thing is it doesn't look like a convergent series - and even if it is I have no idea how to show it - this $\sin$ function inside means I can't use any convergence test (at least I think so). So how to reason about $K_n$?
UPDATE: by absolute convergence od $a_n$ I of course mean that $\sum |a_n|$ is convergent
 A: Call $b_k=\frac{3k^3-2k}{k^3-7}$. This sequence eventually increases towards the limit $3$.
Let us work with $\sum_{k=1}^{n}b_ka_k\sin(k)$. Let us apply summation by parts such that $a_k\sin(k)$ is the factor that gets summed and $b_k$ the factor that gets differentiated ($b_{k+1}-b_k$).  
Call $S_n:=\sum_{k=1}^{n}a_k\sin(k)$. We get 
$$\begin{align}\left|\sum_{k=1}^{n}b_ka_k\sin(k)\right|&=\left|b_{n+1}S_{n+1}-b_1S_1-\sum_{k=1}^{n}(b_{k+1}-b_k)S_{k+1}\right|\\&\leq |b_{n+1}|\cdot|S_{n+1}|+|b_1|\cdot|S_1|+\sum_{k=1}^{n}(b_{k+1}-b_k)\cdot|S_{k+1}|\end{align}$$
Maybe add a constant to the bound to account for the first few terms in this $b_k$ is not monotonic.
But $S_n$ is absolutely convergent because $$\sum_{k=m}^{n}|a_k\sin(k)|\leq\sum_{k=m}^{n}|a_k|<\epsilon,\ \ \ \ \ \ \text{  for $m,n$ large.}$$
Therefore $|S_k|<C$ for some constant $C$. Using this above we get 
$$\begin{align}\left|\sum_{k=1}^{n}b_ka_k\sin(k)\right|&\leq C\left(|b_{n+1}|+|b_1|+\sum_{k=1}^{n}(b_{k+1}-b_k)\right)\\&\leq C\left(4+4+b_{n+1}-b_1\right)\\&\leq12C\end{align}$$
This means that the summation in your problem remains bounded. Since $\ln(n^2+1)\to\infty$  then the quotient tends to zero.
