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Math Help - Help in Interval of convergence problems!

  1. #1
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    Post Help in Interval of convergence problems!

    Q1) I need help in finding the open interval of convergence for the series..

    --- [(k!) (x^k)] / (k^k) ........when k goes from 1 to infinity..Note: do not examine convergence at the endpoint of the interval.


    Q2) Find all the values of the x for which the series converges!

    [(x+2)^2k ] / (2^k) using the ratio test...

    I solved the function using ratio test, but I did not know how to proceed after then...
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    Quote Originally Posted by Vedicmaths View Post
    Q1) I need help in finding the open interval of convergence for the series..

    --- [(k!) (x^k)] / (k^k) ........when k goes from 1 to infinity..Note: do not examine convergence at the endpoint of the interval.
    [snip]
    Using the ratio test you should be able to get

    |x| < \frac{1}{L} where L = \lim_{k \rightarrow \infty} \frac{k^k}{(k+1)^k}.

    One way of finding this limit is to note that

    \frac{k^k}{(k+1)^k} = \left( \frac{k}{k+1} \right)^k = \left(1 - \frac{1}{k+1} \right)^k = \left(1 + \frac{(-1)}{k+1} \right)^k.

    The limit is now seen to be a standard form ......
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    Quote Originally Posted by Vedicmaths View Post
    [snip]
    Q2) Find all the values of the x for which the series converges!

    [(x+2)^2k ] / (2^k) using the ratio test...

    I solved the function using ratio test, but I did not know how to proceed after then...
    Can you get to \lim_{k \rightarrow \infty} \left| \frac{(x+2)^{k+1}}{2}\right| < 1?

    So you need |x+2| < 1 otherwise the limit 'blows up'. Now you need to check what happens when |x+2| = 1, that is, when x = -1 and x = -3 .....
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    I got (x+1)^2 / 2....but I guess it won't change the meaning of what you were explaining. so does that mean for the value of x = -1 and -3 the series converges because of the fact that we need to show that |x+2| < 1?
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    Quote Originally Posted by Vedicmaths View Post
    I got (x+1)^2 / 2....but I guess it won't change the meaning of what you were explaining. so does that mean for the value of x = -1 and -3 the series converges because of the fact that we need to show that |x+2| < 1?
    Actually the series will converge when |x + 2| = 1 because \lim_{k \rightarrow \infty} \left| \frac{(\pm1)^{k+1}}{2}\right| = \frac{1}{2} < 1.
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    Post First one!

    thanks for the help for the second part..

    In the first, I am getting (x) (k^k) / (k+1)^k...will that change everything??
    And the one you showed below, I think that will lead us to the exponent form e^x right??

    thanks for the help Sir!
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    Quote Originally Posted by Vedicmaths View Post
    thanks for the help for the second part..

    In the first, I am getting (x) (k^k) / (k+1)^k...will that change everything??
    And the one you showed below, I think that will lead us to the exponent form e^x right??

    thanks for the help Sir!
    \frac{\frac{(k+1)! \, x^{k+1}}{(k+1)^{k+1}}}{\frac{k! \, x^k}{k^k}} = \frac{(k+1)! \, x^{k+1} \, k^k}{k! \, x^k \, (k+1)^{k+1}} = \frac{(k+1) \, x \, k^k}{(k+1)^{k+1}} = \frac{x \, k^k}{(k+1)^k}.

    Using the ratio test you therefore require

    \lim_{k \rightarrow \infty} \left| \frac{x \, k^k}{(k+1)^k} \right| < 1  \Rightarrow |x| < \frac{1}{\lim_{k \rightarrow \infty} \frac{k^k}{(k+1)^k}}

    which is essentially a less pleasant looking version of what I gave in post #2.

    The value of the limit is e^{-1} = \frac{1}{e}.
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  8. #8
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    wow!

    That was awesome!

    Thanks a lot!
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