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Question Determine if the series converges or diverges $$\sum_{n=1}^\infty \frac{n^2}{n^3+3} $$ I tried the nth term divergence test and got 0, which is inconclusive. I also tried comparison test and ratio test. For the ratio test, I got L=1. This looks easy but I just cannot figure out if the series converges or diverges. Any help is much appreciated.

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    $\begingroup$ You can try the limit comparison test with $\sum{\frac{1}{n}}$ $\endgroup$ – saulspatz Feb 17 '18 at 3:21
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    $\begingroup$ $\frac{n^2}{n^3+3}$ is asymptotically equivalent to $\frac{1}{n}$, so the series diverges as saulspatz mentioned!! $\endgroup$ – Chinnapparaj R Feb 17 '18 at 3:25
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    $\begingroup$ $n^2/(n^3+3)$ is a rational function since both the numerator and denominator are polynomials. A few notes for rational functions: the ratio and root test are always inconclusive for sums of rational functions (just determine what happens for $ 1/n^p$. Also the behavior for rational functions is determined by the leading terms of the numerator and denominator. In your example, the series behaves like $n^2/n^3$. The intuition is that the only terms that matter when $n$ is large are the terms with the largest exponents. $\endgroup$ – JavaMan Feb 17 '18 at 3:33
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Since $$\frac{n^2}{n^3+3}\geq\frac{1}{n+3}, $$ it diverges.

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$\dfrac{n^{2}}{n^{3}+3}\geq\dfrac{n^{2}}{n^{3}+3n^{3}}=\dfrac{1}{4}\dfrac{1}{n}$ and $\displaystyle\sum\dfrac{1}{n}=\infty$.

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Since $$\int_{1}^{\infty}\frac{x^2}{x^3+3}dx$$ diverges, the corresponding series does as well by the Integral Test since the integrand is continuous, positive, and decreasing.

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