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Math Help - Complete Interval of Convergence Question(s)

  1. #1
    Super Member Aryth's Avatar
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    Complete Interval of Convergence Question(s)

    Well, I just need a little help in these... There are parts that I can't completely understand:

    Find the complete interval of convergence for each series

    1. \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n}

    For this one... All I could do was set up the limit for the ratio test:

    L = \lim_{n\to \infty} \left| \frac{(-1)^{n + 2}(x + 5)^{2n + 2}}{(2n + 2)} * \frac{2n}{(-1)^{n + 1}(x + 5)^{2n}}\right|

    2. \sum_{n = 0}^{\infty} \frac{\ln{(n + 1)}x^n}{(n + 1)}

    For this one, I did a bit of work... But I need to know something:

    L = \lim_{n\to \infty} \left| \frac{ln(n + 2)x^{n + 1}}{(n + 2)} * \frac{(n + 1)}{ln(n + 1)x^n}\right|

    = \lim_{n\to \infty} \left| \frac{ln(n+2)(n+1)x}{ln(n+1)(n + 2)}\right|

    = |x|\lim_{n\to \infty} \left| \frac{(ln(n + 2))^{n + 1}}{(ln(n + 1))^{n + 2}}\right|

    This supposedly equals:

    L = |x|

    But I don't really know how the limit gets to 1.

    Any help is appreciated.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Aryth View Post
    Well, I just need a little help in these... There are parts that I can't completely understand:

    Find the complete interval of convergence for each series

    1. \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n}

    For this one... All I could do was set up the limit for the ratio test:

    L = \lim_{n\to \infty} \left| \frac{(-1)^{n + 2}(x + 5)^{2n + 2}}{(2n + 2)} * \frac{2n}{(-1)^{n + 1}(x + 5)^{2n}}\right|

    2. \sum_{n = 0}^{\infty} \frac{\ln{(n + 1)}x^n}{(n + 1)}

    For this one, I did a bit of work... But I need to know something:

    L = \lim_{n\to \infty} \left| \frac{ln(n + 2)x^{n + 1}}{(n + 2)} * \frac{(n + 1)}{ln(n + 1)x^n}\right|..
    = \lim_{n\to \infty} \left| \frac{ln(n+2)(n+1)x}{ln(n+1)(n + 2)}\right|

    = |x|\lim_{n\to \infty} \left| \frac{(ln(n + 2))^{n + 1}}{(ln(n + 1))^{n + 2}}\right|

    This supposedly equals:

    L = |x|

    But I don't really know how the limit gets to 1.

    Any help is appreciated.
    For the first one you should have \bigg|\frac{(x+5)^{2n+1}}{2n+1}\cdot\frac{2n}{(x+5  )^{2n}}\bigg|<1...therefore \lim_{n \to {\infty}}\bigg|\frac{2n(x+5)}{2n+1}\bigg|<1 so |x+5|<1\Rightarrow{-6\leq{x}\leq{-4}}..I got the inclusion by testing those points for convergence in the original series
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  3. #3
    Super Member Aryth's Avatar
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    Thanks that helped quite a bit.

    I'm going to go ahead and finish that one and see if the rest of it is right:

    \text{Int. of Conv} = (-6, -4)

    For x = -6:

    \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(-1)^{2n}}{2n}

    = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n}

    \text{First few terms:}\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - ...

    The alternating series converges because:

    1. |a_{n + 1}| < |a_n|

    2. \lim_{n\to \infty} a_n = 0

    Therefore \sum_{n=1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n} converges at x = -6.

    For x = -4:

    This would lead to the same series, meaning that the sum converges for x = -4 as well.

    \text{Complete Interval of Convergence} = [-6, -4]

    Is that it?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Aryth View Post
    Thanks that helped quite a bit.

    I'm going to go ahead and finish that one and see if the rest of it is right:

    \text{Int. of Conv} = (-6, -4)

    For x = -6:

    \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(-1)^{2n}}{2n}

    = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n}

    \text{First few terms:}\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - ...

    The alternating series converges because:

    1. |a_{n + 1}| < |a_n|

    2. \lim_{n\to \infty} a_n = 0

    Therefore \sum_{n=1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n} converges at x = -6.

    For x = -4:

    This would lead to the same series, meaning that the sum converges for x = -4 as well.

    \text{Complete Interval of Convergence} = [-6, -4]

    Is that it?
    great job!
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Aryth View Post
    Well, I just need a little help in these... There are parts that I can't completely understand:

    Find the complete interval of convergence for each series

    1. \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}(x + 5)^{2n}}{2n}

    For this one... All I could do was set up the limit for the ratio test:

    L = \lim_{n\to \infty} \left| \frac{(-1)^{n + 2}(x + 5)^{2n + 2}}{(2n + 2)} * \frac{2n}{(-1)^{n + 1}(x + 5)^{2n}}\right|

    2. \sum_{n = 0}^{\infty} \frac{\ln{(n + 1)}x^n}{(n + 1)}

    For this one, I did a bit of work... But I need to know something:

    L = \lim_{n\to \infty} \left| \frac{ln(n + 2)x^{n + 1}}{(n + 2)} * \frac{(n + 1)}{ln(n + 1)x^n}\right|

    = \lim_{n\to \infty} \left| \frac{ln(n+2)(n+1)x}{ln(n+1)(n + 2)}\right|

    = |x|\lim_{n\to \infty} \left| \frac{(ln(n + 2))^{n + 1}}{(ln(n + 1))^{n + 2}}\right|

    This supposedly equals:

    L = |x|

    But I don't really know how the limit gets to 1.

    Any help is appreciated.
    For the second one you have \lim_{n \to {\infty}}\bigg|\frac{ln(n+2)x^{n+1}}{n+2}\cdot\fra  c{n+1}{x^{n}ln(n+1)}\bigg|<1...so |x|<1 the way I solved this was by looking at the limit as n gets infinitely large the +1 won't matter...so I extracted teh x and called the limit one...and test for convergence at x=-1,x=1 and I will crique you
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  6. #6
    Super Member Aryth's Avatar
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    These series are quite confusing, our teacher never told us the behavior of series with natural logarithms, for example:

    When x = -1 it gives an alternating series, supposedly it converges, but I can't prove it necessarily since the first term of it is 0 and the second term is -\frac{ln(2)}{2}. Of course only the absolute value of it matters, but the terms in the sequence decrease only when n \geq 3... Does that still imply convergence?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Aryth View Post
    These series are quite confusing, our teacher never told us the behavior of series with natural logarithms, for example:

    When x = -1 it gives an alternating series, supposedly it converges, but I can't prove it necessarily since the first term of it is 0 and the second term is -\frac{ln(2)}{2}. Of course only the absolute value of it matters, but the terms in the sequence decrease only when n \geq 3... Does that still imply convergence?
    Yes it does...lets say the series for n\geq{3} sums to some value denoted b(we know it sums to soem arbitrary value since as you proved the series converges for n\geq{3})? then the real sum of the entire series is a_0+a_2+a_3+b which is obviously convergent
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  8. #8
    Super Member Aryth's Avatar
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    Oh yeah... >_>

    Haha.

    Anyway, when x = 1. All I did was compare it to the harmonic series, which is less than the series we have, and since the harmonic series diverges, then our series diverges as well.

    Giving us:

    \text{Complete Interval of Convergence} = [-1, 1)

    That right?
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Aryth View Post
    Oh yeah... >_>

    Haha.

    Anyway, when x = 1. All I did was compare it to the harmonic series, which is less than the series we have, and since the harmonic series diverges, then our series diverges as well.

    Giving us:

    \text{Complete Interval of Convergence} = [-1, 1)

    That right?
    an easier way is this \sum_{n=0}^{\infty}\frac{ln(n+1)}{n+1}...convergence can be tested this way \int_1^{\infty}\frac{ln(x+1)}{x+1}dx=[ln(x+1)]^2\bigg|_1^{\infty}=\infty therefore it is divergent...and we can use this test becuase a_n is monotonically decreasing...
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  10. #10
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    Quote Originally Posted by Aryth View Post
    Oh yeah... >_>

    Haha.

    Anyway, when x = 1. All I did was compare it to the harmonic series, which is less than the series we have, and since the harmonic series diverges, then our series diverges as well.

    Giving us:

    \text{Complete Interval of Convergence} = [-1, 1)

    That right?
    Yes.
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  11. #11
    Super Member Aryth's Avatar
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    Thanks guys. The help was much needed and much appreciated.
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