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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? – Pete L. Clark Jan 23 '12 at 0:34
Pete: Actually, $1/z$ may well be a number... Which brings us back to this, I believe. :-) – Did Jan 23 '12 at 0:40
@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.) – anon Jan 23 '12 at 0:57
The right formula is $1/L=\lim_{n\to\infty} |a_{n+1}/a_n|$. In your case $a_n=\log(n)$. – emiliocba Jan 23 '12 at 1:29
@Emir: Can you find the radius of converge for $\sum_n z^n$? – JavaMan Jan 23 '12 at 2:14

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.) – Pete L. Clark Jan 23 '12 at 3:04

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