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Let $\sum_2^\infty a_nx^n$ be a power series. Find the radius of convergence when $\lim \limits_{n \to \infty} \frac {a_n}{n^3}$ = 1.

I've tried using root test but that gets messy, can't find a way to use ratio test.

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It is not that messy the root test. As $\lim a_n/n^3 = 1$ you have for sufficiently large $n$ that $1/2 \le a_n/n^3 \le 3/2$ so $n^3/2 \le a_n \le 3n^3/2$. And the $n$-th root of both sides of the inequality is seen easily to approach $1$.

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  • $\begingroup$ Okay great. So R=1 then? $\endgroup$ – Ronique Hossain Mar 2 '15 at 23:39
  • $\begingroup$ Yes, the radius of convergence is $1$. $\endgroup$ – quid Mar 2 '15 at 23:52
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hint: compare it with the series $\displaystyle \sum_{n=1}^\infty n^3x^n$ whose radius of convergence is $1$.

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For all natural $\;n$' s big enough, we have that

$$\frac{n^3}2<a_n<\frac{3n^3}2\;\;\;and\;\;\;a_n\ge 0$$

( just apply the definition of limit with $\;\epsilon=\frac12\;$) , and from here

$$\sqrt[n]\frac{n^3}2\le\sqrt[n]{a_n}\le\sqrt[n]\frac{3n^3}2$$

Apply now the squezze theorem to get $\;\sqrt[n]{a_n}\xrightarrow[n\to\infty]{}1\;$

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