$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}$ 
Given the limit:
$$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \alpha$$
Find the value of $\alpha$


I know the series does not converge (it is equivalent to the harmonic series. Correct me, please, if I am wrong). 
Does it let me apply the following:
$$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \lim_{n\to+\infty}\frac{2n}{n\log(n)(n^2-2n-3)}$$
Or not?
Thank you in advance.

Note that
$$\frac{2k}{k^2-2k-3}=\frac{2k}{(k+1)(k-3)}= \frac{1}{2}\left(\frac{1}{k+1} +\frac{3}{k-3} \right)$$
So, with $H_n = \sum_{k=1}^{n}\frac{1}{n}$
$$\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \frac{1}{2}\sum_{k=4}^{n}\left(\frac{1}{k+1} +\frac{3}{k-3} \right)\leq \frac{1}{2}H_{n+1} + \frac{3}{2}H_{n+1} = 2H_{n+1}\leq 2(\ln{(n+1)}+1)$$
If follows:
$$0\leq\frac{1}{n\ln(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}\leq2\frac{\ln{(n+1)}+1}{n\ln(n)}\stackrel{n\rightarrow\infty}{\longrightarrow}0$$
