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Thread: 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

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

    and

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

    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

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

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

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

    RonL
    Last edited by CaptainBlack; Apr 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 $\displaystyle n \to \infty$ the general term goes to $\displaystyle - \frac 14 \ne 0$
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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

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

    /quote]

    Consider:

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

    So the limit of the general term as $\displaystyle n \to \infnty$ is $\displaystyle -1/4$, 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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    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
    $\displaystyle (1/2)^n$ and $\displaystyle (3/4)^{n + 1}$ both go to zero as $\displaystyle n \to \infty$. 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
    $\displaystyle \sum sin^{2}\frac{1}{n}$

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

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



    RonL


    RonL
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