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Math Help - Convergence series ...

  1. #1
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    Post Convergence series ...

    Hi all!

    How would you do these questions?

    a) Find the range of values of x for convergence of series:

     x + \frac{2^4x^2}{2!}+\frac{3^4x^3}{3!}+\frac{4^4x^4}{  4!} + ...


    b) Investigate convergence of series

     \frac{1}{1.2}+\frac{1}{2.2^2}+\frac{1}{3.2^3}+\fra  c{1}{4.2^4}+ ...

    Thank you! :-)
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  2. #2
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    Quote Originally Posted by Enita View Post
    Hi all!

    How would you do these questions?

    a) Find the range of values of x for convergence of series:

     x + \frac{2^4x^2}{2!}+\frac{3^4x^3}{3!}+\frac{4^4x^4}{  4!} + ...
    The series is,
    \sum_{n=1}^{\infty} \frac{n^4 x^n}{n!}
    Find the limit of,
    \frac{a_{n+1}}{a_n}=\left|\frac{(n+1)^4}{(n+1)!} \cdot \frac{n!}{n^4}\right|=\left|\frac{n!}{n!(n+1)} \cdot \frac{(n+1)^4}{n^4}\right|=\frac{(n+1)^3}{n^4}
    Open,
    \frac{n^3+3n^2+3n+1}{n^4}
    The limit converges to zero.
    Thus, this series only converges for all x.
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  3. #3
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    Quote Originally Posted by Enita View Post

    b) Investigate convergence of series

     \frac{1}{1.2}+\frac{1}{2.2^2}+\frac{1}{3.2^3}+\fra  c{1}{4.2^4}+ ...
    You can see that,
    0\leq \frac{1}{1\cdot 2}+\frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+... \leq \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...
    This sequence is bounded by two convergent sequences.
    Because,
    \frac{1}{2}+\frac{1}{2^2}+...
    Is a geometric series and it converges.

    This is mine 43th Post!!!
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  4. #4
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    thanks theperfecthacker!
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