Question

please help is this absolutely or cond convergent sigma (-1)^n-1 arctan(1/n)/n^2

Answer 1

The term of the absolute series is \(|\arctan(1/n)|/n^2<\frac{\pi/2}{n^2}\). But the series \(\sum 1/n^2\) is convergent (since it is a p-series with p=2), so the absolute series converges. Then the series is absolutely convergent.

Answer 2

how did you come up with pi/2/x^2

Answer 3

the range of the arctan function is between -pi/2 and pi/2

Answer 4

and why did you put over x^2

Answer 5

well I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place.

Answer 6

so would you always compare it to pi/2 divided by denominator given in the problem

Answer 7

for tan

Answer 8

I wouldn't say always. I just see the opportunity that if we compare with pi/2 then I can have some conclusion. It is possible for a problem to have an arctan bu we don't compare with the pi/2.

Answer 9

and if there isnt anything in the denominator of the given problem you would compare it to pi/2

Answer 10

No, I won't if there is nothing on the bottom, I can compare arctan(1/n) < pi/2. But the series \(\sum pi/2\) is divergent. So comparing with pi/2 doesn't give me any conclusion.

Answer 11

okay what would you comoare arcsin (1/n) to
and 1-cos1/n to

Answer 12

the term of the series only arcsin(1/n) ?

Answer 13

yah

Answer 14

Just to make sure. So your series is
\(\sum_{n=1}^\infty \arcsin(1/n)\)?
I can't think how to do this right away...

Answer 15

yah

Answer 16

ok, arcsin is an increasing function and for positive x we have
x > sin x
Apply the arcsin
arcsin x > x
It follows that
arcsin(1/n) > 1/n
But the harmonic series \(\sum 1/n\) is divergent
Hence \(\sum \arcsin(1/n)\) is divergent as well.

Answer 17

what if it was just sin

Answer 18

sin what?

Answer 19

if instead of arc sin it was sin would you still make the same comparison

Answer 20

you mean sin(1/n) ?

Answer 21

yah

Answer 22

well we can't compare to 1/n since 1/n > sin(1/n)
So I don't know the answer yet.

Answer 23

oh okay

Answer 24

also if you were given sigma 1-cos(1/n) what would you compare it to

Answer 25

That is actually a nice problem. I'll post it as a new question so everybody can give a response (BTW it is convergent)