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Math Help - Alt Series test

  1. #1
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    Alt Series test

    \sum^ \infty_ {n = 1} \frac{(-1)^n}{ln(n+1)}

    How do I use the Alt series test on this summation? I'm pretty sure that the function is > 0, but I'm not sure if its decreasing.
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  2. #2
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    Quote Originally Posted by saiyanmx89 View Post
    \sum^ \infty_ {n = 1} \frac{(-1)^n}{ln(n+1)}

    How do I use the Alt series test on this summation? I'm pretty sure that the function is > 0, but I'm not sure if its decreasing.
    Is \ln(n+1) increasing?

    CB
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  3. #3
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    Well, you can't get a negative number with ln(anything) as long as there are no negatives outside the function, so it is indeed positive. And it is decreasing, because as n increases, the denominator increases, so the values are getting smaller and smaller.
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  4. #4
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    once I took the limit of the absolute value of the function, I found it to be divergent. Am I correct? I hope so....
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  5. #5
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    Quote Originally Posted by saiyanmx89 View Post
    once I took the limit of the absolute value of the function, I found it to be divergent. Am I correct? I hope so....
    Please state the alternating series test (here is some help: Alternating series test - Wikipedia, the free encyclopedia)

    Now look at the previous posts again.
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  6. #6
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    1) \frac{1}{ln(n+1)}> 0
    2) \frac{1}{ln(n+1)} is decreasing
    3) \lim_{n \rightarrow \infty} \frac{1}{ln(n+1)} = \lim_{n \rightarrow \infty} (n+1) = \infty

    By Alt Series Test, \sum^ \infty _{n=1} \frac{(-1)^n}{ln(n+1)} is divergent.
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  7. #7
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    Quote Originally Posted by saiyanmx89 View Post
    1) \frac{1}{ln(n+1)}> 0
    2) \frac{1}{ln(n+1)} is decreasing
    3) \lim_{n \rightarrow \infty} \frac{1}{ln(n+1)} = \lim_{n \rightarrow \infty} (n+1) = \infty Mr F says: How do you figure this?

    By Alt Series Test, \sum^ \infty _{n=1} \frac{(-1)^n}{ln(n+1)} is divergent.
    The limit in 3) is equal to zero. The
    = \lim_{n \rightarrow \infty} (n+1)
    bit is totally wrong.
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