Let $R > 0$ be the radius of convergence of a power series $\sum\limits_{n=0}^\infty a_n(z-a)^n $. Show that the radius of convergence of the power series $\sum\limits_{n=0}^\infty \frac{a_n} {n+1}(z -a)^{n+1}$ is greater than or equal to $R$.
I have one approach that uses Cauchy Hadamard theorem to conclude that $\lim_{n \to \infty} |\frac{a_n}{n+1}|^{1/n} \leq |a_n|^{1/n}$. Hence $R_2\geq R_1$
Is there a simpler solution?

  • $\begingroup$ It seems hard to find something simpler than these two lines… $\endgroup$
    – Bernard
    Sep 17, 2017 at 18:49
  • $\begingroup$ But I'm not sure if this is the correct way $\endgroup$
    – john doe
    Sep 17, 2017 at 18:50
  • 1
    $\begingroup$ This is one way. You also can prove that if the given series converges, the new series also converges. It should use normal convergence of the given series on every closed disk with radius $<R$. $\endgroup$
    – Bernard
    Sep 17, 2017 at 18:59
  • $\begingroup$ @Bernard but I also have to show that radius of convergence is $\geq $ original R $\endgroup$
    – john doe
    Sep 17, 2017 at 19:02
  • $\begingroup$ If it is $\ge r$ for all $r<R$, it is $\ge R$. Isn't it clear to you? $\endgroup$
    – Bernard
    Sep 17, 2017 at 19:04

1 Answer 1


Hints: $$\sum\limits_{n=0}^\infty\frac{a_n}{n+1}(z-a)^{n+1} = (z-a)\sum\limits_{n=0}^\infty\frac{a_n}{n+1}(z-a)^n,$$ $$\left|\frac{a_n}{n+1}\right|\le|a_n|,$$ and use direct comparison.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .