Evaluation of product of $n$ terms of series , If $n\rightarrow \infty$ 
If $a_{1},a_{2},a_{3},.....,a_{n}$ are n terms of series such that $$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}\;,n\geq 1, n\in \mathbb{N}$$
Then $\displaystyle n^4\lim_{n\rightarrow \infty}\prod^{n}_{r=1}a_{r} = $

$\bf{My\; Try::}$ Given $$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}=2+\frac{4}{n}$$
We can Write it as $$\frac{n+1}{a_{n+1}}-\frac{n}{a_{n}} = \frac{1}{a_{n}}+2+\frac{4}{n}$$
Now Put $n=1,2,3,4,.......n$ and adding, We get
$$\frac{n+1}{a_{n+1}}-\frac{1}{a_{n}} = \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+......+\frac{1}{a_{n}}\right)+2n+4\left(\frac{1}{1}+\frac{1}{2}+.....+\frac{1}{n}\right)$$
Now How can I solve after that, Help me
Thanks

HINT : 
We can write 
 $$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}$$
as
$$b_{n+1}-b_n=n+2$$
where
$$b_n=\frac{\binom n2}{a_n}$$
