the radius of convergence of the power series

**sinichko** · November 15th 2010, 01:36 PM

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 $z=1,-1, i$

**Prove It** · November 15th 2010, 02:34 PM

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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