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Math Help - ratio, root tests

  1. #1
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    Oct 2008
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    ratio, root tests

    Match each of the following with the correct statement.
    A. The series is absolutely convergent.
    C. The series converges, but is not absolutely convergent.
    D. The series diverges.

    \sum_{n=1}^\infty<br />
    (-1)^{n+1}\frac{(6 + n) 2^n}{(n^2) 3^{2 n}}
    <br />
\sum_{n=1}^\infty<br />
    \left( \frac{ n^{7} }{ 7 - 10 n^{4} } \right)^n

    \sum_{n=1}^\infty<br />
    \frac{ (- 3 n)^n }{ n^{5 n} }

    \sum_{n=1}^\infty<br />
    \frac{ (n+5)^n }{5^{n^2}}

    \sum_{n=1}^\infty<br />
    \frac{ (n+5)^n }{ 5^n }

    \sum_{n=1}^\infty<br />
    (-1)^n n^{- 1}\ln ( n + 3 )

    Can someone please help me?

    -thanks
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  2. #2
    Banned
    Joined
    Nov 2008
    Posts
    82

    GOOD OLD Fashion ratio test

    Quote Originally Posted by fastcarslaugh View Post
    Match each of the following with the correct statement.
    A. The series is absolutely convergent.
    C. The series converges, but is not absolutely convergent.
    D. The series diverges.

    \sum_{n=1}^\infty<br />
(-1)^{n+1}\frac{(6 + n) 2^n}{(n^2) 3^{2 n}}
    <br />
\sum_{n=1}^\infty<br />
\left( \frac{ n^{7} }{ 7 - 10 n^{4} } \right)^n

    \sum_{n=1}^\infty<br />
\frac{ (- 3 n)^n }{ n^{5 n} }

    \sum_{n=1}^\infty<br />
\frac{ (n+5)^n }{5^{n^2}}

    \sum_{n=1}^\infty<br />
\frac{ (n+5)^n }{ 5^n }

    \sum_{n=1}^\infty<br />
(-1)^n n^{- 1}\ln ( n + 3 )

    Can someone please help me?

    -thanks
    hey mate,

    there are numerous ways to test converge of sums of infinite series, however the most common of which is to use the ratio test,
    Ratio Test :
    Given an infinite sum of the form
    S = sum (a(n), n = 0..inf)
    then if L < 1, the series converges, L > 1 the series diverges and finally if L = 1 the test fails and nothing can be stated (by the Ratio test) about its convergence,

    where

    L = lim (n--> inf) |a(n+1)/a(n)|

    As an example I will do one of your questions where
    a(n) = (-1)^n * (1/n)*(1/ln(n+3))
    then a(n+1) = (-1)^(n+1) * (1/(n+1))*(1/ln(n + 4))
    thus,
    a(n+1)/a(n) = (-1)^(n+1) * (1/(n+1))*(1/ln(n + 4)) / ((-1)^n * (1/n)*(1/ln(n+3)))
    = (-1)*(n/(n+1))*(ln(n+3)/ln(n+5))
    thus,
    |a(n+1)/a(n)| = (n/(n+1))*(ln(n+3)/ln(n+5))

    we now need to evaluate the limit,
    L = lim (n --> inf) (n/(n+1))*(ln(n+3)/ln(n+5))
    note it is easily observed that lim (n--> inf) (n/(n+1)) = 1 then L becomes
    L = lim (n --> inf) ln(n+3)/ln(n+5)
    = infinity/infinity
    Hence employ L'hopitals Rule
    L = lim (n--> inf) d/dn (ln(n+3))/ d/dn (ln(n+5))
    L = lim (n--> inf) (1/(n+3))/(1/(n+5))
    L = lim (n--> inf) (n+5)/(n+3) = 5/3 > 1
    Thus the series diverges.

    Hope this was informative and I didnt go overboard with the description.

    Regards,

    David

    ps - forgot to mention that for something to be absolutely convergent then the sum( |a(n)| , n = 0..inf ) must converge, in order to determine this, simply take the absolute value of the integrand of summation a(n) and apply the Ratio test as prescribed.
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  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    35
    So far, I got

    D for the 1st problem
    D for the 2nd problem
    A for the 3rd problem
    D for the 5th probem

    and i'm lost after that
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  4. #4
    Junior Member
    Joined
    Oct 2008
    Posts
    35
    I finally got it

    thank you everyone who looked at my problem.
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