Show $\frac1{k-1}+\frac 6k-\frac{7}{k+1}=\frac{8k-6}{k(k^2-1)}$, deduce $\sum\limits_{k = 2}^n\frac{4k-3}{k(k^2-1)}$

Question

Prove that if $k > 1$, then

$$ \dfrac{1}{k - 1} + \dfrac{6}{k} - \dfrac{7}{k + 1} = \dfrac{8k - 6}{k(k^2 - 1)}$$

Hence simplify

$$ \sum\limits_{k = 2}^n \dfrac{4k - 3}{k(k^2 -1)} $$

My Work

$$ \dfrac{1}{k - 1} + \dfrac{6}{k} - \dfrac{7}{k + 1} = \dfrac{8k - 6}{k(k^2 - 1)} = \dfrac{7k - 6}{k(k - 1)} - \dfrac{7}{k + 1}$$

$$= \dfrac{(7k - 6)(k + 1) - 7(k(k - 1))}{k(k - 1)(k + 1)}$$

$$= \dfrac{8k - 6}{k(k^2 - 1)} = 2 \left[ \dfrac{4k - 3}{k(k^2 -1)} \right]$$

$$\therefore \sum_{k = 2}^n \dfrac{4k - 3}{k(k^2 - 1)} = \dfrac{1}{2} \sum_{k = 2}^n \left( \dfrac{1}{k -1} + \dfrac{6}{k} - \dfrac{7}{k + 1} \right)$$

$$= \dfrac{1}{2} \left( \sum_{k = 2}^n \dfrac{1}{k -1} + 6 \sum_{k = 2}^n \dfrac{1}{k} - 7 \sum_{k = 2}^n \dfrac{1}{k + 1} \right)$$

$$= \dfrac{1}{2} \left( \sum_{k = 1}^{n - 1} \dfrac{1}{k - 1} + 6 \sum_{k = 2}^n \dfrac{1}{k} - 7 \sum_{k = 3}^{n + 1} \dfrac{1}{k + 1} \right)$$

I would appreciate it if people could please take the time to review my work and explain how I should proceed.

Answer 1

\begin{align*} \sum\limits_{k = 2}^n \dfrac{4k - 3}{k(k^2 -1)} & = \frac{1}{2}\sum\limits_{k = 2}^n \dfrac{8k - 6}{k(k^2 -1)}\\ &=\frac{1}{2}\sum_{k = 2}^n \left( \dfrac{1}{k - 1} + \dfrac{6}{k} - \dfrac{7}{k + 1} \right)\\ &=\frac{1}{2}\sum_{k = 2}^n \left( \dfrac{1}{k - 1} + \dfrac{6}{k} - \dfrac{6}{k + 1}-\dfrac{1}{k+1} \right)\\ &=\frac{1}{2}\left[\sum_{k = 2}^n \left( \dfrac{1}{k - 1} -\dfrac{1}{k+1}\right) +6\sum_{k = 2}^n \left( \dfrac{1}{k} - \dfrac{1}{k + 1}\right)\right]\\ &=\frac{1}{2}\left[\left(1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right)+6\left(\frac{1}{2}-\frac{1}{n+1}\right)\right] \end{align*} In the last step, I am using the idea of a telescopic sum. I hope you can take it from here.