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Math Help - [SOLVED] Series (Ratio test)

  1. #1
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    [SOLVED] Series (Ratio test)

    Hey there.. i've done the ration test for the series at the attachment..
    i've got the answer 1/7 where the series is convergent..
    can someone confirm my answer..?
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  2. #2
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    Quote Originally Posted by nameck View Post
    Hey there.. i've done the ration test for the series at the attachment..
    i've got the answer 1/7 where the series is convergent..
    can someone confirm my answer..?
    Pretty straight forward, isn't it? a_n= \frac{n^3}{7^n} so a_{n+1}= \frac{(n+1)^3}{7^{n+1}} and the ratio is \frac{a_{n+1}}{a_n}= \frac{(n+1)^3}{7^{n+1}}\frac{7^n}{n^3} = \frac{(n+1)^3}{n^3}\frac{7^n}{7^{n+1}}= \frac{n^3+ 3n^2+ 3n+ 1}{n^3}\frac{1}{7}= \left(1+ \frac{3}{n}+ \frac{3}{n^2}+ \frac{1}{n^3}\right)\left(\frac{1}{7}\right).

    As n goes to infinity, those fractions go to 0 so the limit (1)(\frac{1}{7})< 1. Yes, the series converges.
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    THANKS HallsofIvy
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