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- anonymous

does this converge or diverge sigma n=1 to infinity 1/n^2 tan 1/n

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- anonymous

does this converge or diverge sigma n=1 to infinity 1/n^2 tan 1/n

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- anonymous

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

- anonymous

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

- 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 ^_^

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- anonymous

wow, great answer! medal'd

- anonymous

lol, thank you ^_^

- anonymous

but please check if there's any mistake :)

- anonymous

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

- anonymous

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

- anonymous

k

- anonymous

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

- 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 :)

- anonymous

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

- anonymous

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

- 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 ^_^

- anonymous

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

- anonymous

k ty u for help

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

thank you, likewise :)

- anonymous

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

- anonymous

are you an expert
at writng

- anonymous

want to do a google doc
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