Question

Does the series [absolute value of (sin n)]/(n^2+1) converge or diverge?

Answer 1

\[\sum_{1}^{\infty} \frac{ \left| \sin n \right| }{ n ^{2} + 1 }\]

Answer 2

i forgot how to do this things
i guess you do this by comparison test

Answer 3

just a try not sure \[|\sin n|\leq1 \Longrightarrow \frac{|\sin n|}{n^2+1}\leq \frac{1}{n^2+1}\]
take the limit as n goes to oo
\[\lim_{n\to\infty}\frac{1}{n^2+1}=0\]
hence
\[\lim_{n\to\infty}\frac{|\sin n|}{n^2+1}=0\]
as a result the sum converges

Answer 4

but i'm not sure this correct lol :)

Answer 5

did you do comparison test?

Answer 6

Thanks!! :)

Answer 7

I was thinking of the comparison test, but I wasn't quite sure how to do it

Answer 8

welcome :)
check your notes on how is this done
and see if that's what i did^^

Answer 9

@surjithayer

Answer 10

None of the examples from my notes are with absolute values, but the make up series is of the highest power, so I think the regular series should be \[\frac{ 1 }{ n ^{2} }\]

Answer 11

well i know 1/(n^2+1) converges since 1/n^2 does converge as well

Answer 12

So, you're supposed to leave the absolute value of sine alone?

Answer 13

hmm not sure what you meant by that?

Answer 14

Uhmm during the comparison test, what are you supposed to do with 'sin n'?

Answer 15

bring help @myininaya

Answer 16

well we are supposed to compare the giving series with a different series that we are familiar with and that's what i did above

Answer 17

if that series that we know converges so is our series
if not then the series deos not converge this is what comparison test is about

Answer 18

So would it start like this:
\[\frac{ \left| \sin n \right| }{ n ^{2}+1 } \le \frac{ \left| \sin n \right| }{ n ^{2} }\]

Answer 19

refer to this page, there are some examples of comparison test

Answer 20

no see my work above

Answer 21

But that isn't quite like what my teacher showed us

Answer 22

well i don't know what your teacher showed you
but what i did above is correct!
i'm just not sure if i need more steps to say that it is convergent

Answer 23

if you look at the link i posted they are doing the same thing

Answer 24

I don't see a link

Answer 25

http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx

Answer 26

i thought i posted it

Answer 27

haha xD

Answer 28

\[\large \frac{|\sin n|}{n^2}\] how would you prove this convergent or not

Answer 29

that's what your were thought in class?

Answer 30

Well, we were only taught with numerators of 1 and that's why I'm not sure what to do with

Answer 31

to do with sin n *

Answer 32

well look above
i have 1/ (n^2+1) and that's what you needed
this series converge just line 1/n^2

Answer 33

But what about \[\left| \sin n \right|\] ?

Answer 34

hehe okay you didn't got what i did
by definition of sin we have:
\[|\sin n|\leq1\] yes

Answer 35

?

Answer 36

agree?

Answer 37

yes

Answer 38

okay then i multiplied the inequality by \[\frac{1}{n^2+1}\] which is positive so no change in the inequaltiy
i get
\[\large \frac{|\sin n|}{n^2+1}\leq\frac{1}{n^2+1}\]
no get the form i needed
\[\Large a_n\leq b_n\]
since\[\Large b_n\] converges
then \[\large a_n\] also converges

Answer 39

now*

Answer 40

does it make sense now?

Answer 41

Yeahh! thanks!

Answer 42

welcome!