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Math Help - Ratio Test

  1. #1
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    Ratio Test

    a_n =  \frac{ (6 n + 6) 6^{n + 3} }{ 4^{n} }<br />


    Can someone please explain the steps of solving this

    -Thanks
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  2. #2
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    If \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1, then \sum a_n is absolutely convergent.

    So: \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left| {\color{red}a_{n+1}} \cdot {\color{blue}\frac{1}{a_n}} \right|  = \lim_{n \to \infty} \left| {\color{red}\frac{\left(6(n+1) + 6\right)6^{(n+1)+3}}{4^{n+1}}} \cdot {\color{blue}\frac{4^n}{(6n + 6)6^{n+3}}}\right| = \hdots

    See where you can go from here. Post back any specific problems you may have.
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  3. #3
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    I guess I asked a wrong question there


    I already got to that part it's just that I am not sure about canceling the numbers
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  4. #4
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    Well simplifying a bit we get: \lim_{n \to \infty} \left| \frac{\left(6n+ 12\right)6^{n+4}}{4^{n+1}} \cdot \frac{4^n}{(6n + 6)6^{n+3}}\right|

    Make use of the property that: \frac{a^m}{a^n} = a^{m-n}

    So we get: = \lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot 4^{n-(n+1)} \cdot 6^{n+4 - (n+3)}\right| = \lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot \frac{6}{4}\right| = \cdots
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  5. #5
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    So I realized that I was able to simplify this correctly.

    so I pull out 3/2 then \lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot \right|

    so top is approaching infinity faster than the bottom. Does this mean that it's 1?

    So the limit is 3/2 and the ratio test is inconclusive. But will the series converge or diverge by another test?
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  6. #6
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    Quote Originally Posted by fastcarslaugh View Post
    So I realized that I was able to simplify this correctly.

    so I pull out 3/2 then \lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot \right|

    so top is approaching infinity faster than the bottom. Does this mean that it's 1?

    So the limit is 3/2 and the ratio test is inconclusive. But will the series converge or diverge by another test?
    It is not inconclusive, if the ratio test yields a number greater than one it is divergent. If you'd care for further proof of its divergence why not try the n-th term test for divergence?
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  7. #7
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    For some random reason, the program that I use to do my homework would not accept the correct answer. After 37th times, it finally accepted the answer. and I believe that I typed the same correct answer about 34 times.

    anyway thanks for clearing my doubt.
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