$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}$

Question

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 $\alpha$

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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.

Answer 1

HINT

By Stolz-Cesaro

$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}}{n\log n}=$$

$$=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{\frac{2n+2}{(n+1)^2-2(n+1)-3}}{(n+1)\log(n+1)-n\log n}=\lim_{n\to\infty}\frac{\frac{2n+2}{n^2-4}}{\log \left(1+\frac1n\right)^n+\log (n+1)}$$

Answer 2

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: $$0\leq\frac{1}{n\ln(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}\leq2\frac{\ln{(n+1)}+1}{n\ln(n)}\stackrel{n\rightarrow\infty}{\longrightarrow}0$$

Answer 3

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.

$\lim\limits_{n\to+\infty} \frac{1}{n}+\frac{\gamma+\frac{25}{24}}{n\ln(n)}+\frac{\big(-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}\big)}{n\ln(n)}$