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How to find the radius of convergence of the series $\sum z^{n^2}$ . All I know is $\sum a_n(z-z_0)^n$ . My intuitions says that it's ROC should be 1, as it can be seen as the sub series of $\sum z^{n}$ . But how can I prove it.

Thanks and regards!

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    You have edited the question but not added anything. The fact that it is a subseries of $\sum z^n$ is not important. My original answer suggested the comparison test, which applies term by term. Do you see how that applies here? – 2017-10-05
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    Ya it can be seen , as for |z|<1 we have z^(n^2) < z^n and hence it's convergence follows. – 2017-10-05
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    Yes, that is the point I was making. And if $|z| \gt 1...$ – 2017-10-05
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    Certainly if |z|>1,we have z ^(n^2) >z^n which diverges as z^n diverges . – 2017-10-05
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    That makes the outline of the proof. If you run into troubles, that would make a new question. – 2017-10-05
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    Is it enough to get an Upvote. @Ross Millikan please give me one I want to ask more questions. – 2017-10-05
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    I don't know why you would be prohibited from asking questions as you have positive reputation and haven't asked any lately. I am not one of the downvotes, but this question is just a homework problem with no effort shown. The fact that you edited it without adding useful information could earn some more. – 2017-10-05
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    @RossMilkikan I don't know either , I have been blocked by the site to ask anymore questions . I thought that getting Upvotes could help in this case , If you could help in removing those downvotes – 2017-10-06

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Hint: compare with $\sum z^n$.

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    Now I am getting it , the only difference between these two series is their rate of convergence one convergence with faster rate while the other with slower is it so?? – 2017-01-18
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    It is true that the radius of convergence is $1$, but you need to prove it. To prove it converges for $|z| \lt 1$ you can use the comparison test with the series I gave. You need another tool for non-convergence for $|z| \gt 1$ – 2017-01-18