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Find the interval of convergence of $$\sum_{n=1}^\infty \frac{(-1)^{n+1}(x-2)^n}{n2^n} $$

Thank you very much in advance!

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Define $z=x+2$, where does $$\sum_{n=1}^\infty \frac{(-1)^{n+1} z^n}{n 2^n}$$ converge? What can you say about your sum in advance?

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Hint: 1) Try the Alternating series test. This should give you an open interval, where the series converges.

2) Find and argument why it diverges outside the corresponding closed interval (do the arguments even go to zero?)

3) Treat the boundary seperately.

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You can also use the fact that the radius of convergens is equal to:

$$R=\frac{1}{\limsup_{n\rightarrow\infty}{|a_n|}}$$ In this case: $$a_n=\frac{(-1)^{n+1}}{n\cdot2^n}$$ Calculation gives: the series converges whenever $|z|<2$.

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So does that mean, -2 < x < 0? –  Cossette Mar 19 '13 at 10:52
    
If we have just $x^n$ instead of $(x-2)^n$ we would get $-2<x<2$, thus what does that mean for $(x-2)^2$?! –  Adrian57 Mar 19 '13 at 10:58
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