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nth term test

n=1(2n3)(n8+1)2(65n)2(43n2)2\displaystyle\sum\limits_{n=1}^{\infty }{\frac{(2n-3)(n^8+1)^2}{{(6-5n)^2(4-3n^2)^2}}}
What conclusion can be reached by using the nthn^{\text{th}} term test on the series?
Choose 1 answer:
Choose 1 answer: