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? $\endgroup$– DidCommented Jan 21, 2016 at 14:51
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2 Answers
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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.
answered Nov 25, 2015 at 11:07
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$\begingroup$ Shouldn't it be $R=\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}$? $\endgroup$– MOACommented Nov 25, 2015 at 11:14
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$\begingroup$ @Olga, $\frac1R=\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_{n}}$ $\endgroup$ Commented Nov 25, 2015 at 11:16
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$\begingroup$ 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...). $\endgroup$– DidCommented Jan 19, 2016 at 22:22
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$\begingroup$ @Did, true. Should I delete my answer? $\endgroup$ Commented Jan 20, 2016 at 7:29
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One nice characterisation of the radius of convergence $R$ of a power series $\sum_{n\ge 0} a_n x^n$ is given by the Cauchy-Hadamard theorem:
$$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}$$
Try applying this to your two sums.
