Apply the Limit Comparison Test to ∑n=1∞n2+1n4−2n2+1\sum_{n=1}^{\infty} \frac{n^2 + 1}{n^4 - 2n^2 + 1}∑n=1∞n4−2n2+1n2+1 using bn=1n2b_n = \frac{1}{n^2}bn=n21. What is limn→∞anbn\lim_{n \to \infty} \frac{a_n}{b_n}limn→∞bnan?
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∞\infty∞
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2