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?
$\begingroup$
$\endgroup$
7
-
$\begingroup$ It seems hard to find something simpler than these two lines… $\endgroup$– BernardSep 17, 2017 at 18:49
-
$\begingroup$ But I'm not sure if this is the correct way $\endgroup$– john doeSep 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 $<R$. $\endgroup$– BernardSep 17, 2017 at 18:59
-
$\begingroup$ @Bernard but I also have to show that radius of convergence is $\geq $ original R $\endgroup$– john doeSep 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? $\endgroup$– BernardSep 17, 2017 at 19:04
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
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.
answered Sep 17, 2017 at 20:36
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
41.1k4 gold badges45 silver badges89 bronze badges