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How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooarctan(n)/(n^1.2)# ?

Answer 1

William Abdalla

Since

#arctann le pi/2#,

we have

#{arctann}/n^{1.2} le {pi/2}/n^{1.2}#.

Since #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#,

by Comparison Test,

#sum_{n=1}^infty {arctann}/n^{1.2}# also converges.

I hope that this was helpful.

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Answer 2

Christopher Allers

To use the Direct Comparison Test on the infinite series (\sum_{n=1}^{\infty} \frac{\arctan(n)}{n^{1.2}}), consider the series (\sum_{n=1}^{\infty} \frac{1}{n^{1.2}}). Note that (\frac{\arctan(n)}{n^{1.2}}) is always less than or equal to (\frac{1}{n^{1.2}}) since (\arctan(n)) is less than or equal to (\pi/2) for all (n).

Since (\frac{\arctan(n)}{n^{1.2}} \leq \frac{1}{n^{1.2}}) and (\sum_{n=1}^{\infty} \frac{1}{n^{1.2}}) is a convergent p-series with (p > 1), by the Direct Comparison Test, the series (\sum_{n=1}^{\infty} \frac{\arctan(n)}{n^{1.2}}) converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.