How do you use the direct Comparison test on the infinite series $\infty \sum n = 2 \frac{n^{3}}{n^{4} - 1}$ $sum_(n=2)^oon^3/(n^4-1)$ ?

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How do you use the direct Comparison test on the infinite series $\infty \sum n = 2 \frac{n^{3}}{n^{4} - 1}$ $sum_(n=2)^oon^3/(n^4-1)$ ?

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Answer 1 · 2014-09-08T18:26:12

Since $\frac{n^{3}}{n^{4} - 1} \geq \frac{n^{3}}{n^{4}} = \frac{1}{n}$ $n^3/{n^4-1} geq n^3/n^4=1/n$ for all $n \geq 2$ $n geq2$ and $\infty \sum n = 2 \frac{1}{n}$ $sum_{n=2}^infty1/n$ is a harmonic series, which is known to be divergent, we may conclude that $\infty \sum n = 2 \frac{n^{3}}{n^{4} - 1}$ $sum_{n=2}^inftyn^3/{n^4-1}$ also diverges by Direct Comparison Test.

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How do you use the direct Comparison test on the infinite series $\infty \sum n = 2 \frac{n^{3}}{n^{4} - 1}$ $sum_(n=2)^oon^3/(n^4-1)$ ?

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How do you use the direct Comparison test on the infinite series sum_(n=2)^oon^3/(n^4-1)∞∑n=2n3n4−1sum_(n=2)^oon^3/(n^4-1) ?

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How do you use the direct Comparison test on the infinite series $\infty \sum n = 2 \frac{n^{3}}{n^{4} - 1}$ $sum_(n=2)^oon^3/(n^4-1)$ ?