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Infinite Seriesmedium
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Apply the Limit Comparison Test to ∑n=1∞n2+3n2n4−n2+1\sum_{n=1}^{\infty} \frac{n^2 + 3n}{2n^4 - n^2 + 1}∑n=1∞​2n4−n2+1n2+3n​ by comparing with bn=1n2b_n = \frac{1}{n^2}bn​=n21​. What is lim⁡n→∞anbn\lim_{n \to \infty} \frac{a_n}{b_n}limn→∞​bn​an​​?