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Math Help - Series Problem

  1. #1
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    Arrow Series Problem

    Hello, I need help with this series:

    The instructions are: Find all the values of x such that the given series would converge. The series is convergent
    from , left end included (enter Y or N):
    to , right end included (enter Y or N):
    Now, I used the ratio test but I got suck with the square root of n, could someone show the steps to solve the series with the ratio test? I know after the values are found you put in each value of x in the series then solve using a test like the limit comparison test to see if it converges so that you know if it is included for the values.
    Thanks,
    Matt
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  2. #2
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    \lim_{n \to \infty} \left|\frac{10^{n+1}x^{n+1}}{\sqrt{n+1}+2} \cdot \frac{\sqrt{n}+2}{10^n x^n} \right| < 1

    |10x| \lim_{n \to \infty} \frac{\sqrt{n}+2}{\sqrt{n+1}+2} < 1

    |10x| \lim_{n \to \infty} \frac{1 + \frac{2}{\sqrt{n}}}{\sqrt{1+\frac{1}{\sqrt{n}}}+\f  rac{2}{\sqrt{n}}} < 1

    |10x| \cdot 1 < 1

    interval of convergence is

    -\frac{1}{10} < x \leq \frac{1}{10}
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  3. #3
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    Thank you! But, what happened to the (-1)^n from the numerator from the original series? Wouldn't it be just -1, how does it go away?\

    Edit: Oh wait, is it because (-1)^infinity is indeterminate?
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  4. #4
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    absolute value ?
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  5. #5
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    Question

    Also, what method did you use to find out if the endpoints were included for the interval of convergence?
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  6. #6
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    Quote Originally Posted by skeeter View Post
    absolute value ?
    haha, I forgot it was there. Thanks.
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  7. #7
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    Quote Originally Posted by matt3D View Post
    Also, what method did you use to find out if the endpoints were included for the interval of convergence?
    sub each endpoint for x in the original nth term.

    for x = 1/10, you get an alternating series whose nth term goes to 0.

    for x = -1/10, the series diverges ... why?
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  8. #8
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    Is it because p=1/2 which is less than 1 for -1/10, so that makes it divergent? The p-series is from the integral test right?
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  9. #9
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    Well after plugging it in and simplifying, yep:

    \sum_{n = 1}^{\infty} \frac{(-1)^n \cdot 10^n \cdot \displaystyle \frac{1}{(-10)^n}}{\sqrt{n + 2}} = \sum_{n = 1}^{\infty} \frac{(-10)^n \cdot \displaystyle \frac{1}{(-10)^n}}{\sqrt{n+2}} = \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n+2}} = \sum_{n=3}^{\infty} \frac{1}{\sqrt{n}}
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  10. #10
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    Cool, I didn't know you could take the 2 from the denominator and add it to n in the limit to make it 3...
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