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Math Help - the radius of convergence of the power series

  1. #1
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    the radius of convergence of the power series

    Please help me with this one...i don't really understand how can i to solve it, and what the nth coefficient here?

    Show that the radius of convergence of the power series:

    \sum_{n=1}^{\infty }\frac{(-1)^{n}}{n}z^{n(n+1)}

    is 1 and discuss convergence for z=1,-1, i
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  2. #2
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    Try the ratio test...

    \displaystyle a_{n} = \frac{(-1)^nz^{n(n+1)}}{n}, so \displaystyle a_{n+1} = \frac{(-1)^{n+1}z^{(n+1)(n+2)}}{n+1}.


    The series will be convergent when \displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1

    \displaystyle \lim_{n \to \infty}\left|\frac{\frac{(-1)^{n+1}z^{(n+1)(n+2)}}{n+1}}{\frac{(-1)^nz^{n(n+1)}}{n}}\right|<1

    \displaystyle \lim_{n \to \infty}\left|\frac{n(-1)^{n+1}z^{(n+1)(n+2)}}{(n+1)(-1)^nz^{n(n+1)}}\right| < 1

    \displaystyle \lim_{n \to \infty}\left|\frac{n\,z^{(n+1)(n+2)}}{(n+1)z^{n(n+  1)}}\right|<1

    \displaystyle \lim_{n \to \infty}\left|\frac{n\,z^2}{n+1}\right|<1

    \displaystyle |z^2|\lim_{n \to \infty}\frac{n}{n+1} < 1

    \displaystyle |z|^2\lim_{n \to \infty}\frac{1}{1 + \frac{1}{n}} < 1

    \displaystyle |z|^2\cdot 1 < 1

    \displaystyle |z|^2 < 1

    \displaystyle |z| < 1.

    So the series is convergent when \displaystyle |z| < 1 and divergent when \displaystyle |z| > 1. You will need to do some further analysis to determine the convergence/divergence when \displaystyle |z| = 1.
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