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Thread: converge or diverge

  1. #1
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    converge or diverge

    Can you guys check my work to see if I did it right

    the function is [(n^2)(2^n-1)]/(-5^n)

    so what i did was use the nth root test and got (n^(2/n))/10..(is the lim of n^(2/n) equal to 1??..i just used my calculator for this..)

    so the limit of that would be 1/10 <1 therefore converge

    am i right?
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  2. #2
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    Quote Originally Posted by johntuan View Post
    the function is [(n^2)(2^n-1)]/(-5^n)
    There are some real notational problems to this question.

    $\displaystyle -5^n$ is a negative number. $\displaystyle \sum {\frac{1}{{ - 5^n }} = - \sum {\frac{1}{{5^n }}} } $.

    Do you mean $\displaystyle (-5)^n$?


    Is it $\displaystyle 2^n-1$ or $\displaystyle 2^{n-1}?$
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  3. #3
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    Quote Originally Posted by Plato View Post
    There are some real notational problems to this question.

    $\displaystyle -5^n$ is a negative number. $\displaystyle \sum {\frac{1}{{ - 5^n }} = - \sum {\frac{1}{{5^n }}} } $.

    Do you mean $\displaystyle (-5)^n$?


    Is it $\displaystyle 2^n-1$ or $\displaystyle 2^{n-1}$
    sorry about the notation problem

    for the (-5), I thought it didnt matter because when using the nth-root test, its the absolute value that's using.

    so yeah i meant $\displaystyle (-5)^n$

    and $\displaystyle 2^{n-1}?$ is the correct one.

    sorry for the confusion
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  4. #4
    Senior Member Twig's Avatar
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    Hi

    I assume you mean $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}} $

    $\displaystyle |\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}|\leq \frac{n^{2}(2^{n})}{5^{n}} $

    So this series is absolute convergent.
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  5. #5
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    Quote Originally Posted by Twig View Post
    Hi

    I assume you mean $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}} $

    $\displaystyle |\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}|\leq \frac{n^{2}(2^{n})}{5^{n}} $

    So this series is absolute convergent.
    no theres no (-1)^n in the numerator and (-5^n) on the denominator
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