## anonymous 5 years ago does this converge or diverge sigma n=1 to infinity 1/n^2 tan 1/n

1. anonymous

hmm, this is a tough one. going to have to bust out my calc resources =)

2. anonymous

check out this resource: http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx

3. anonymous

you can use the squeeze theorem here , now we have the following:$\sum_{n=1}^{\infty} \frac{\tan(\frac{1}{n})}{n^2}$ So by squeeze theorem we have : $\frac{-\pi}{2} < \tan \frac{1}{n} < \frac{\pi}{2}$ divide by n^2 both sides and you'll get : $\lim_{x \rightarrow \infty}\frac{-\pi}{2n^2} <\lim_{x \rightarrow \infty} \tan \frac{1}{n} <\lim_{x \rightarrow \infty} \frac{\pi}{2n^2}$ where : $\lim_{x \rightarrow \infty} \frac{-\pi}{2n^2} = \lim_{x \rightarrow \infty} \frac{\pi}{2n^2} = \lim_{x \rightarrow \infty} \tan \frac{1}{n} = 0$ where series = a finite number = 0, then it converges, correct me if I'm wrong ^_^

4. anonymous

5. anonymous

lol, thank you ^_^

6. anonymous

but please check if there's any mistake :)

7. anonymous

can u please explain why tan of 1/n is b/w -pi/2 and pi/2

8. anonymous

first, graph tan (n) on paper, you'll notice that tan (whateve angle it has here) is between -pi/2 and pi/2 :)

9. anonymous

k

10. anonymous

if you sketch all tans on the graph, then you'll see different intervals that you can take ^_^

11. anonymous

other thatn -pi/2 and pi/2, depends on the tan that you want to choose, but I chose the simplest of all lol :)

12. anonymous

and how did u got from the inequality sign to the equal sign

13. anonymous

I didn't , I just applied the sqeeze theorem and wrote it down :)

14. anonymous

which says if limit of (a) as x -->infinity = 0 and limit of (b) as x--> infinity = 0 then limit of (c, the one that's in the middle) = 0 ^_^

15. anonymous

-pi/2 = a and pi/2 = b, so c = tan(1/n) ^_^

16. anonymous

k ty u for help

17. anonymous

u guys are brilliant in math ...........;)

18. anonymous

np ^_^ lol sath, I wasn't at all, but I worked hard to reach this level :)

19. anonymous

haha u all are deserving student .....best of luck..

20. anonymous

thank you, likewise :)

21. anonymous

am the student of medicine but only becauz of some iq question am here.....

22. anonymous

are you an expert at writng

23. anonymous

want to do a google doc \