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If $a_{k}=\frac{k^4-17k^2+16}{k^4-8k^2+16}$ real

number for $5≤k\in N$

Then find :

$\lim_{n\to +\infty} a_{5}a_{6}a_{7}...a_{n}$

My try :

$k^4-17k^2+16=(k^2-1)(k^2-16)$

And

$k^4-8k^2+16=(k-4)^2$

But how I complete

Answer is $\frac{1}{14}$

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Let $a_k = p_k/q_k$, where $p_k = k^4 - 17k^2 + 16 = (k-4)(k-1)(k+1)(k+4)$, and $q_k = k^4 - 8k^2 + 16 = (k-2)(k-2)(k+2)(k+2)$. Then $$\prod_{k=5}^n a_k = \prod_{k=5}^n \frac{(k-4)(k-1)(k+1)(k+4)}{(k-2)(k-2)(k+2)(k+2)}.$$ This is a telescoping product; for instance, $$\prod_{k=5}^n \frac{k-4}{k-2} = \frac{\prod_{k=1}^{n-4} k}{\prod_{k=3}^{n-2} k} = \frac{1 \cdot 2}{(n-3)(n-2)}.$$ Treat the other fractions similarly and you will find your answer.

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