What is the radius of convergence of $\sum_{n= 1 }^{n=\infty}a_n x^n$ ? Where $a_n $ is the nth prime. I know it can not be bigger than one because at $x =1$ series is just sum of all primes which is divergent. I also tried nth prime bounds but not got much. Any hint regarding this will be helpful. Thank you.
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2?? I think you missed something...and anyway $$\sqrt[n]{p_n}=1$$ so the raidus of convergence is 1... – DonAntonio Jul 29 '20 at 18:37
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How to show this limit? – CHOUDHARY bhim sen Jul 29 '20 at 18:42
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As you know $p_n \sim n \log n$ for $n$ large enough . So looking at large $n$ you should get, $n \le p_n \le n^2$. Now take $$\lim_{n \to \infty} n^{1/n} \le \lim_{n \to \infty} {p_n}^{1/n} \le \lim_{n \to \infty} {n}^{2/n}$$
Thus the radius of convergence is 1
Brozovic
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