Question

Determine whether the series is absolutely convergent, conditionally convergent, or divergent

Answer 1

\[\sum_{n=1}^{\infty}(-1)^n \frac{ \arctan(n) }{ n^8 }\]

Answer 2

Are you familiar with the alternating series test? If you can show that \(\dfrac{\arctan n}{n^8}\) approaches 0 and is a decreasing sequence, the series converges.

To establish absolute convergence, you need to show that \(\sum \dfrac{\arctan n}{n^8}\) converges.

Answer 3

i was wondering if the p series applied here, but i wasnt sure if i would be able to use 1/n^8

Answer 4

im sorry, i was working on another problem and didnt see you posted a response

Answer 5

hi x

Answer 6

hello there :)

Answer 7

i was asking about the p series cuz i noticed if i could use \[\frac{ 1 }{ n^8 }\] and the p series, \[8>1\]= convergent

Answer 8

You could, but first you must show that \(\dfrac{1}{n^8}\le\dfrac{\arctan n}{n^8}\).

Answer 9

idk how to show that lol thats why i posted this

Answer 10

don't know what they are actually... lol

Answer 11

isnt that as \[n^8\] approaches infinity, arctan aproaches \[\frac{ \pi }{ 2 }\]?

Answer 12

and that means convergent at pi/2?

Answer 13

I'm really tired can't even think lol
you meant \[\lim_{n\to \infty }\arctan(n)=\pi/2\]

Answer 14

that's right don't if that will help you though

Answer 15

yes lol and its ok

Answer 16

no worries, i know the answer is absolutely convergent, i was just wondering how we come to that realization

Answer 17

im sticking with the p series, even tho i dont know how to show its smaller than a_n

Answer 18

oh, i haven't done such thing before i mean abs converg
conditional converg?!!

Answer 19

yea,.. its totally ridiculous

Answer 20

haha anyways i will check this out later, now i'm really not me lol

Answer 21

no worries, im gonna close this question! have a good sleep

Answer 22

I'm half asleep as i speak :)

Answer 23

you too take care

Answer 24

To show that \(\dfrac{1}{n^8}\le\dfrac{\arctan n}{n^8}\) is the same as showing that \(1\le \arctan n\) for some large enough \(N>n\ge1\). You have that \(\arctan 1=\dfrac{\pi}{4}\approx0.758398\), but \(\arctan2\approx1.10715\), so \(N=2\) is sufficient.

So you have that for \(n\ge2\),
\[\arctan n\ge1~~\implies~~\frac{\arctan n}{n^8}\ge\frac{1}{n^8}\]
and since \(\sum\dfrac{1}{n^8}\) converges, so must \(\sum\dfrac{\arctan n}{n^8}\). The fact that the inequality \(\arctan n\ge1\) is true for \(n\ge2\) doesn't make a difference when you want to show that \(\sum\dfrac{\arctan n}{n^8}\) converges.

This is because
\[\sum_{n=1}^\infty \frac{\arctan n}{n^8}=\frac{\pi}{4}+\sum_{n=2}^\infty \frac{\arctan n}{n^8}\ge\frac{\pi}{4}+\sum_{n=1}^\infty\frac{1}{n^8}\]
\(\dfrac{\pi}{4}\) is a finite number, and adding a finite term to a convergent series does not change the fact that the series converges.

Answer 25

Thank you so much!