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Suppose $\sum \limits_{n=0}^{\infty} a_n x^n$ has radius of convergence R. What is the radius of convergence of $\sum \limits_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1}$?

How do I solve this without using power series integration?

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  • $\begingroup$ Which characterizations of the radius of convergence of a series do you know? $\endgroup$
    – Did
    Jan 21 '16 at 14:51
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Given $R=\lim\limits_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right|=\lim\limits_{n\to\infty} \left|\frac{a_{n-1}}{a_{n}}\right|$

Also, let $$\sum \limits_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1}=\sum \limits_{n=1}^{\infty} \frac{a_{n-1} x^{n}}{n}=\sum \limits_{n=1}^{\infty} b_{n} x^{n}$$

where $b_n=\frac{a_{n-1}}{n}$

$R_1=\lim\limits_{n\to\infty} \left|\frac{b_n}{b_{n+1}}\right|=\lim\limits_{n\to\infty} \left|\frac{a_{n-1}}{n}\times\frac{n+1}{a_{n}}\right|=\lim\limits_{n\to\infty} \left|\frac{n+1}{n}\right|\times \lim\limits_{n\to\infty} \left|\frac{a_{n-1}}{a_{n}}\right|$. You know where to go from here.

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  • $\begingroup$ Shouldn't it be $R=\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}$? $\endgroup$
    – MOA
    Nov 25 '15 at 11:14
  • $\begingroup$ @Olga, $\frac1R=\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_{n}}$ $\endgroup$ Nov 25 '15 at 11:16
  • $\begingroup$ Ah yeah, thanks! $\endgroup$
    – MOA
    Nov 25 '15 at 11:17
  • $\begingroup$ This does not solve the exercise since it may well happen that the ratios $\left|\frac{a_n}{a_{n+1}}\right|$ have no limit (these could even be undefined for infinitely many $n$s...). $\endgroup$
    – Did
    Jan 19 '16 at 22:22
  • $\begingroup$ @Did, true. Should I delete my answer? $\endgroup$ Jan 20 '16 at 7:29

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