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If $n^3 < |a_n| < n^4$ find the radius of convergence for $\sum_{n=2}^\infty a_nx^n$

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Could someone explain how he got inequality (1)? Theorem 4.1 stated that a power series converges if $|x| <R$.

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The radius of convergence of the series $\sum_n b_nx^n$ is the supremum of the set $$\{t\geqslant 0, \sup_{n\geqslant 0}|b_n|t^n<+\infty\}.$$ Since $$\{t\geqslant 0, \sup_{n\geqslant 0}n^4t^n<+\infty\}\subset \{t\geqslant 0, \sup_{n\geqslant 0}|a_n|t^n<+\infty\} \subset \{t\geqslant 0, \sup_{n\geqslant 0}n^3t^n<+\infty\},$$ the conclusion follows.

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  • $\begingroup$ How does that use the comparison theorem? $\endgroup$ – user144464 May 11 '14 at 11:55
  • $\begingroup$ It's actually not needed if we use this definition of the radius of convergence. If we look at the non-negative $t$ such that the series $\sum_n b_nt^n$ is convergent, then indeed the comparison test is used: if $0\leqslant a_n\leqslant b_n$ and $\sum_n b_nt^n<\infty$ then $\sum_n a_nt^n<\infty$. $\endgroup$ – Davide Giraudo May 11 '14 at 12:04

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