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Math Help - Alternating series

  1. #1
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    Alternating series

    I'm trying to show that this series diverges.

    <br />
\sum_{n=1}^{\infty} (-1)^n\frac{1+4^n}{1+3^n}<br />

    I'm really drawing a blank here. I know there is a quick way to show divergence but I'm not sure how. Any help appreciated.
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  2. #2
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    Quote Originally Posted by Shananay View Post
    I'm trying to show that this series diverges.

    <br />
\sum_{n=1}^{\infty} (-1)^n\frac{1+4^n}{1+3^n}<br />

    I'm really drawing a blank here. I know there is a quick way to show divergence but I'm not sure how. Any help appreciated.
    If \color{blue}\lim_{n\to\infty} |a_n| \neq 0 then \color{blue}\lim_{n\to\infty} a_n D.N.E
    Hence, \color{blue}\sum an diverges by the nth term test (The test for divergence).
    your \color{blue}a_n is \color{blue}(-1)^n\frac{1+4^n}{1+3^n}.
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  3. #3
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    Quote Originally Posted by General View Post
    If \color{blue}\lim_{n\to\infty} |a_n| \neq 0 then \color{blue}\lim_{n\to\infty} a_n D.N.E
    Hence, \color{blue}\sum an diverges by the nth term test (The test for divergence).
    your \color{blue}a_n is \color{blue}(-1)^n\frac{1+4^n}{1+3^n}.
    Thank you, that's kind of what I was thinking to do but I realized I didn't even know how to evaluate that limit. I know it's not equal to zero but how to I show that? I was getting \frac{\infty}{\infty}, but using L'Hopital, I was still getting the same answer.
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  4. #4
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    Quote Originally Posted by Shananay View Post
    Thank you, that's kind of what I was thinking to do but I realized I didn't even know how to evaluate that limit. I know it's not equal to zero but how to I show that? I was getting \frac{\infty}{\infty}, but using L'Hopital, I was still getting the same answer.
    When you do L`Hospital`s Rule you will face:
    \color{blue}\lim_{n\to\infty} \frac{4^n ln(4)}{3^n ln(3)}=\frac{ln(4)}{ln(3)}\lim_{n\to\infty} (\frac{4}{3})^n=?

    Remember that:
    1- \color{blue} \lim_{n\to\infty} L^n = \infty if \color{blue}|L|>1
    2- \color{blue} \lim_{n\to\infty} L^n = 0 if \color{blue}|L|<1
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  5. #5
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    Quote Originally Posted by General View Post
    When you do L`Hospital`s Rule you will face:
    \color{blue}\lim_{n\to\infty} \frac{4^n ln(4)}{3^n ln(3)}=\frac{ln(4)}{ln(3)}\lim_{n\to\infty} (\frac{4}{3})^n=?

    Remember that:
    1- \color{blue} \lim_{n\to\infty} L^n = \infty if \color{blue}|L|>1
    2- \color{blue} \lim_{n\to\infty} L^n = 0 if \color{blue}|L|<1
    Perfect, thank you
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