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Convergence tests challenge

If, using the direct comparison test, I compare each of the series listed below to   n=1 n22n2+3~~\displaystyle\sum\limits_{n=1}^{\infty }~{\frac{n^2}{2n^2+3}}\,, what can I conclude?
I. n=1n22n2+4\qquad\displaystyle\sum\limits_{n=1}^{\infty }{\frac{n^2}{2n^2+4}}
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II. n=1 2n22n2+3\qquad\displaystyle\sum\limits_{n=1}^{\infty }~{\frac{2n^2}{2n^2+3}}
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III. n=1 n2n2+3\qquad\displaystyle\sum\limits_{n=1}^{\infty }~{\frac{n}{2n^2+3}}
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Choose 1 answer: