1. watchmath

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.

2. anonymous

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

3. watchmath

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

4. anonymous

and why did you put over x^2

5. watchmath

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

6. anonymous

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

7. anonymous

for tan

8. watchmath

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.

9. anonymous

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

10. watchmath

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.

11. anonymous

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

12. watchmath

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

13. anonymous

yah

14. watchmath

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...

15. anonymous

yah

16. watchmath

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.

17. anonymous

what if it was just sin

18. watchmath

sin what?

19. anonymous

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

20. watchmath

you mean sin(1/n) ?

21. anonymous

yah

22. watchmath

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

23. anonymous

oh okay

24. anonymous

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

25. watchmath

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