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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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    @Emir: Can you find the radius of converge for $\sum_n z^n$?2012-01-23

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Using the Cauchy-Hadamard theorem:

$\frac{1}R=\limsup_{n\to\infty} \left|\frac{1}{\ln(n)}\right|^\frac{1}{n} = 1$

Then $R=1$.

EDIT: Alternatively, the same result can be obtained using the (less general) ratio test, since: $\frac{1}R=\lim_{n\to\infty} \left|\frac{\ln(n+1)}{\ln(n)}\right| = 1$

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    Well, this is certainly correct, but the OP asked about the correctness of *his* argument. So this is not a maximally helpful answer. (And do you really think that someone who thinks a radius of convergence might be $\frac{1}{z}$ knows about the limit superior? To me it seems unlikely.)2012-01-23