$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}$ 
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.
 A: 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:
$$0\leq\frac{1}{n\ln(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}\leq2\frac{\ln{(n+1)}+1}{n\ln(n)}\stackrel{n\rightarrow\infty}{\longrightarrow}0$$
A: HINT
By Stolz-Cesaro
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}}{n\log n}=$$
$$=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{\frac{2n+2}{(n+1)^2-2(n+1)-3}}{(n+1)\log(n+1)-n\log n}=\lim_{n\to\infty}\frac{\frac{2n+2}{n^2-4}}{\log \left(1+\frac1n\right)^n+\log (n+1)}$$
A: From the $\lim\limits_{n\to+\infty} \frac{1}{n\log(n)}\sum\limits_{k=4}^{n}\frac{2k}{k^2-2k-3}$ first dealing with the sum equals to the followings: 
$\sum\limits_{k=4}^{n}\frac{2k}{k^2-2k-3}=\sum\limits_{k=4}^{n}\frac{2k}{(k+1)(k-3)}=\frac{1}{2}\sum\limits_{k=4}^{n}\big(\frac{k}{k-3}-\frac{k}{k+1}\big)=\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{k+3}{k}-\frac{k+3}{k+4}\big)=\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{3}{k}-\frac{1}{k+4}\big)$
Forming further: 
$\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{3}{k}-\frac{1}{k+4}\big)=\sum\limits_{k=1}^{n}\frac{k+4+2}{k(k+4)}=\sum\limits_{k=1}^{n}\frac{1}{k}+\sum\limits_{k=1}^{n}\frac{2}{k(k+4)}=\sum\limits_{k=1}^{n}\frac{1}{k}+\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{1}{k}-\frac{1}{k+4}\big)$
$\sum\limits_{k=1}^{n}\big(\frac{1}{k}-\frac{1}{k+4}\big)$ is a harmonic series equal to $\frac{50}{24}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}$
Finally 
$\lim\limits_{n\to+\infty} \frac{1}{n\ln(n)}\big(\sum\limits_{k=1}^{n}\frac{1}{k}+\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{1}{k}-\frac{1}{k+4}\big)\big)=\lim\limits_{n\to+\infty} \frac{1}{n\ln(n)}\big(\gamma+\ln(n)+\frac{25}{24}+\frac{1}{2}\big(-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}\big)\big)$ 
where $\gamma$ is the Euler-Macheroni constant.
$\lim\limits_{n\to+\infty} \frac{1}{n}+\frac{\gamma+\frac{25}{24}}{n\ln(n)}+\frac{\big(-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}\big)}{n\ln(n)}$
