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What is the R.O.C. of the following power series:

$$\sum_{n\geq2}\frac{z^{n}}{\ln(n)}\qquad?$$ Here is my attempt:

$$\lim_{n\rightarrow\infty}\left|\frac {z^{n+1}\ln(n)}{\ln(n+1)z^{n}}\right|=\lim_{n\rightarrow\infty}\left|\frac{z\ln(n)}{\ln(n+1)}\right|=z$$ so the R.O.C. = $\frac {1}{z}$. Is this right?

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    Well, here's a first check: the radius of convergence of a power series is a number (or possibly $+\infty$), right? Is $\frac{1}{z}$ a number?2012-01-23
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    Pete: Actually, $1/z$ may well be a number... Which brings us back to [this](http://math.stackexchange.com/q/96420/6179), I believe. :-)2012-01-23
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    Unless I messed up the ratio test, the way I learned it is that the radius of convergence = $\frac{1}{L}$ where L is the limit obtained from the ratio test.2012-01-23
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    @Emir: That rule you cite is not quite correct. How does it make any sense that the range of possible $z$ values actually *depends* on the variable $z$? (Answer: It doesn't make any sense.)2012-01-23
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    The right formula is $1/L=\lim_{n\to\infty} |a_{n+1}/a_n|$. In your case $a_n=\log(n)$.2012-01-23
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    @Emir: Can you find the radius of converge for $\sum_n z^n$?2012-01-23

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