How do you use the direct Comparison test on the infinite series $\infty \sum n = 1 \frac{arctan (n)}{n^{1.2}}$ $sum_(n=1)^ooarctan(n)/(n^1.2)$ ?

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How do you use the direct Comparison test on the infinite series $\infty \sum n = 1 \frac{arctan (n)}{n^{1.2}}$ $sum_(n=1)^ooarctan(n)/(n^1.2)$ ?

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Answer 1 · 2014-10-15T17:50:47

Since

$arctan n \leq \frac{π}{2}$ $arctann le pi/2$ ,

we have

$\frac{arctan n}{n^{1.2}} \leq \frac{\frac{π}{2}}{n^{1.2}}$ ${arctann}/n^{1.2} le {pi/2}/n^{1.2}$ .

Since
$\infty \sum n = 1 \frac{\frac{π}{2}}{n^{1.2}} = \frac{π}{2} \infty \sum n = 1 \frac{1}{n^{1.2}}$ $sum_{n=1}^infty {pi/2}/n^{1.2}=pi/2 sum_{n=1}^infty 1/n^{1.2}$

is a convergent p-series with $p = 1.2 > 1$ $p=1.2 > 1$ ,

by Comparison Test,

$\infty \sum n = 1 \frac{arctan n}{n^{1.2}}$ $sum_{n=1}^infty {arctann}/n^{1.2}$ also converges.

I hope that this was helpful.

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How do you use the direct Comparison test on the infinite series $\infty \sum n = 1 \frac{arctan (n)}{n^{1.2}}$ $sum_(n=1)^ooarctan(n)/(n^1.2)$ ?

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How do you use the direct Comparison test on the infinite series ∞∑n=1arctan(n)n1.2sum_(n=1)^ooarctan(n)/(n^1.2) ?

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How do you use the direct Comparison test on the infinite series $\infty \sum n = 1 \frac{arctan (n)}{n^{1.2}}$ $sum_(n=1)^ooarctan(n)/(n^1.2)$ ?