radius of convergence of integral of power series

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?

• Which characterizations of the radius of convergence of a series do you know?
– Did
Commented Jan 21, 2016 at 14:51

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.

• Shouldn't it be $R=\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}$?
– MOA
Commented Nov 25, 2015 at 11:14
• @Olga, $\frac1R=\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_{n}}$ Commented Nov 25, 2015 at 11:16
• Ah yeah, thanks!
– MOA
Commented Nov 25, 2015 at 11:17
• 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...).
– Did
Commented Jan 19, 2016 at 22:22
• @Did, true. Should I delete my answer? Commented Jan 20, 2016 at 7:29

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.