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Math Help - Need Help Finding the Integral of Convergence?

  1. #1
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    Need Help Finding the Integral of Convergence?

    hey!

    first off i'm new to this forum

    got a few questions on my homework that I am stuck on. I represent the summation sign with a capital E. n! means n factorial

    1) E (n! x^2n / (3n-1)!)

    2) E (-3/sqrt(n))(x/2)^n

    thanks a bunch if you help me out!
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  2. #2
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    Quote Originally Posted by JhonnyBoi View Post
    hey!

    first off i'm new to this forum

    got a few questions on my homework that I am stuck on. I represent the summation sign with a capital E. n! means n factorial

    1) E (n! x^2n / (3n-1)!)

    2) E (-3/sqrt(n))(x/2)^n

    thanks a bunch if you help me out!
    For the first use the ratio test

    \lim_{ n \to \infty}\left| \frac{a_{n+1}}{a_n} \right|= \lim _{n\to \infty}\left| \frac{x^{2(n+1)}}{[3(n+1)-1]!}\cdot \frac{(3n-1)!}{x^{2n}}\right|=

    \lim_{n \to \infty} \left| \frac{x^{2}}{(3n+2)(3n+1)(3n)} \right|=0

    By the ratio test the radius of convergence is all real numbers....

    Try this test on the 2nd and see what happens
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  3. #3
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    Quote Originally Posted by JhonnyBoi View Post
    hey!

    first off i'm new to this forum

    got a few questions on my homework that I am stuck on. I represent the summation sign with a capital E. n! means n factorial

    1) E (n! x^2n / (3n-1)!)

    2) E (-3/sqrt(n))(x/2)^n

    thanks a bunch if you help me out!
    You mean "interval" of convergence. A fairly standard method of determining radius of convergence is to use the ratio test:
    A series \sum a_n converges if the ratio |a_{n+1}/a_n|< 0.

    For 1, a_n= \frac{n!x^{2n}}{(3n-1)!)} and a_{n+1}= \frac{(n+1)! x^{2(n+1)}{(3(n+1)-1)!}= \frac{(n+1)!x^{2n}x^2}{(3n+2)!}. The ratio is |\frac{(n+1)!x^{2n}x^2}{(3n+2)!}\frac{(3n-1)!}{n!x^{2n}}| = \frac{(n+1)!}{n!}\frac{x^{2n}}{x^{2n}\frac{(3n-1)!}{(3n+2)!}x^2 = \frac{n+1}{(3n+2)(3n+1)(3n)}x^2. Take the limit as n goes to infinity, set that < 1, and solve for x.

    For 2, a_n= \frac{-3}{\sqrt{n}}(x/2)^n, a_{n+1}= \frac{-3}{\sqrt{n+1}}(x/2)^{n+1}. The ratio is |\frac{-3}{\sqrt{n+1}}(x/2)^{n+1}\frac{\sqrt{n+1}}{-3(x/2)^n}| =\frac{\sqrt{n+1}}{\sqrt{n}}(x/2). Again, take the limit as n goes to infinity, set it < 1, and solve for x.
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  4. #4
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    hey

    first off thanks for the help and your right its interval not integral. that was my bust

    Quote Originally Posted by HallsofIvy View Post
    For 2, a_n= \frac{-3}{\sqrt{n}}(x/2)^n, a_{n+1}= \frac{-3}{\sqrt{n+1}}(x/2)^{n+1}. The ratio is |\frac{-3}{\sqrt{n+1}}(x/2)^{n+1}\frac{\sqrt{n+1}}{-3(x/2)^n}| =\frac{\sqrt{n+1}}{\sqrt{n}}(x/2). Again, take the limit as n goes to infinity, set it < 1, and solve for x.


    when you multiply it by the reciprical wouldn't it be (-3/sqrt(n+1))(x/2)^n+1 x (sqrt(n)/-3(x/2)^n) and not what you have

    thanks
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  5. #5
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    bump
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