prove that the series summation from n=1 to infinity from (n+1) /(n+2) z^n converges if lzl<1 and diverges if lzl>1

thanks

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- May 2nd 2010, 05:54 AM #1

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- May 2nd 2010, 06:08 AM #2

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You can either use one of the many ways to find radius of convergence (e.g., root or ratio tests), or you might notice similarity to the geometric series.

For example, $\displaystyle \sum_{n=0}^\infty z^n$ is known to converge for $\displaystyle \lvert z\rvert <1$ and to diverge for $\displaystyle \lvert z\rvert >1$. You could say that since $\displaystyle \frac{1}{2}\leq\frac{n+1}{n+2}\leq1$, then $\displaystyle \frac{1}{2}\sum_{n=0}^\infty z^n\leq\sum_{n=0}^\infty \frac{n+1}{n+2}z^n\leq\sum_{n=0}^\infty z^n$. One of these inequalities shows convergence at $\displaystyle \lvert z\rvert <1$, and the other shows divergence at $\displaystyle \lvert z\rvert >1$ (it’s not hard to figure out which is which).