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Math Help - A sequence problem.

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    A sequence problem.

    Hi, I think I have an idea on how to solve a sequence problem but I could be going badly wrong. To verify if I used the right method I will ask somebody to help me with this question from Multivariable Calculus 6th Edition by James Stewart.

    You have to find if {arctan 2n} converges, and if it converges to find its limit.
    Do I have to find the limit of 0.218n + 0.889? The decimals are approximate so I don't think this could be it.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Undefdisfigure View Post
    Hi, I think I have an idea on how to solve a sequence problem but I could be going badly wrong. To verify if I used the right method I will ask somebody to help me with this question from Multivariable Calculus 6th Edition by James Stewart.

    You have to find if {arctan 2n} converges, and if it converges to find its limit.
    Do I have to find the limit of 0.218n + 0.889? The decimals are approximate so I don't think this could be it.
    Hint: arctan(2n) converges to the same thing arctan(n) does
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    Quote Originally Posted by Jhevon View Post
    Hint: arctan(2n) converges to the same thing arctan(n) does
    I'm having cognitive flashes on how to put this together but I can't quite make it. I'll keep reading my text and maybe I'll get it, I'll put this one aside and return to it. I know your probably reluctant to give me the answer since you gave me a hint.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Undefdisfigure View Post
    I'm having cognitive flashes on how to put this together but I can't quite make it. I'll keep reading my text and maybe I'll get it, I'll put this one aside and return to it. I know your probably reluctant to give me the answer since you gave me a hint.
    we are talking about sequences, right?

    so we want \lim \arctan 2n

    Let u = 2n*

    then \lim \arctan 2n = \lim_{u \to \infty} \arctan u = \frac {\pi}2

    *) the reason why i can substitute a dummy variable here is simply because \lim_{n \to \infty} 2n = \lim_{n \to \infty} n
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    Quote Originally Posted by Jhevon View Post
    we are talking about sequences, right?

    so we want \lim \arctan 2n

    Let u = 2n*

    then \lim \arctan 2n = \lim_{u \to \infty} \arctan u = \frac {\pi}2

    *) the reason why i can substitute a dummy variable here is simply because \lim_{n \to \infty} 2n = \lim_{n \to \infty} n
    You are very intelligent. Thank you. Did you pass Calculus 5? I failed Calculus 3 and I'm taking it over.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Undefdisfigure View Post
    You are very intelligent.
    ...umm, not really

    Thank you.
    you're welcome

    Did you pass Calculus 5?
    yes. but call it adv calc 2. the 5 makes it seem hard and causes people to overestimate my abilities

    I failed Calculus 3 and I'm taking it over.
    sorry to hear that. better luck this time around. things should be easier this time. i guess i can expect to see you here regularly, huh?
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    Quote Originally Posted by Undefdisfigure View Post
    Did you pass Calculus 5?
    What topics would Calculus 5 cover??

    -Dan
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    Quote Originally Posted by topsquark View Post
    What topics would Calculus 5 cover??
    My real multidimensional analysis class covered:
    • topology of \mathbb{R}^n i.e. open,closed,connectness,compactnes,...
    • differenciability in \mathbb{R}^n, Taylor's theorem in \mathbb{R}^n, the general chain rule, Hessian and critical points
    • the implicit function theorem
    • theory of integration on a line, theory of integration in \mathbb{R}^n, Fubini's theorems, chain of variable - Jabocian
    • line integrals, surface integrals, Green's and Divergence theorems
    • Fourier series, convergence of Fourier series, theorems of which functions can be written in Fourier series
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    Quote Originally Posted by Jhevon View Post
    we are talking about sequences, right?

    so we want \lim \arctan 2n

    Let u = 2n*

    then \lim \arctan 2n = \lim_{u \to \infty} \arctan u = \frac {\pi}2

    *) the reason why i can substitute a dummy variable here is simply because \lim_{n \to \infty} 2n = \lim_{n \to \infty} n
    When arctan u approaches infinity, how do you know that it equals PI/2, is it a hard and fast rule I can find in the text?
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    Quote Originally Posted by Undefdisfigure View Post
    When arctan u approaches infinity, how do you know that it equals PI/2, is it a hard and fast rule I can find in the text?
    Do you know what the graph of y = arctan(x) looks like? Your original question implies a background that ought to make the answer 'yes'. Contemplate this graph .......

    Alternatively, contemplate the graph of y = tan(x) .......
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