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Math Help - convergence and differgence

  1. #1
    Member akhayoon's Avatar
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    convergence and differgence

    find out whether this is divergent, c.convergent or abs convergent

    \sum\frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}

    and

    \sum sin^{2}\frac{1}{n}

    so I've basically ruled out the c.convergence..

    but i havent the slightest idea how to start the first one?

    for the second I thought that 1/n is divergent, which means the the whole series would be divergent..?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by akhayoon View Post
    find out whether this is divergent, c.convergent or abs convergent

    \sum\frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}
    Consider:

    \frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}=\frac{4^n[(1/2)^n-1]}{4^{n+1}[(3/4)^{n+1}+1]}=\frac{(1/2)^n-1}{4[(3/4)^{n+1}+1]}

    So the limit of the general term as n \to \infty is -1/4, so the series diverges.

    RonL
    Last edited by CaptainBlack; April 21st 2008 at 10:18 PM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by akhayoon View Post
    might wanna fix that syntax error lol
    he is just using the test for divergence. as n \to \infty the general term goes to - \frac 14 \ne 0
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  4. #4
    Member akhayoon's Avatar
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    [quote=CaptainBlack;133801]
    Quote Originally Posted by akhayoon View Post
    find out whether this is divergent, c.convergent or abs convergent

    \sum\frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}

    /quote]

    Consider:

    \frac{2^{n}-4^{n}}{3^{n+1}+4^{n+1}}=\frac{4^n[(1/2)^n-1]}{4^{n+1}[(3/4)^{n+1}+1]}=\frac{(1/2)^n-1}{4[(3/4)^{n+1}+1]}

    So the limit of the general term as n \to \infnty is -1/4, so the series diverges.

    RonL

    so i don't understand how you reached -1/4...how did you get rid of all the rest..confused
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by akhayoon View Post
    so i don't understand how you reached -1/4...how did you get rid of all the rest..confused
    (1/2)^n and (3/4)^{n + 1} both go to zero as n \to \infty. they are just convergent geometric series. what is left if you make those two zero, is -1/4
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by akhayoon View Post
    \sum sin^{2}\frac{1}{n}

    \sum \left(\sin\left[\frac{1}{n}\right]\right)^2

    converges as \sin(x) \sim x as x \to 0, so you can trap ((\sin(1/n))^2 between positive multiples of (1/x)^2 and we know that \sum_{n=1}^{\infty}1/n^2 converges and so the given series converges by the comparison test (among other means).



    RonL


    RonL
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