3
$\begingroup$

Suppose that there is a power series with coefficients $a_n$ that has a positive radius of convergence. Let $b_n=\sum_{k=0}^{n}a_k$ and show that $$\sum_{n=0}^{\infty}b_nx^n$$ has a positive radius of convergence. I tried using the definition of the radius of convergence $R_a=\frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}}$ but that does not see to help. How would I go about proving this?

$\endgroup$

2 Answers 2

3
$\begingroup$

Since the radius of convergence of $\sum_{n=0}^\infty a_nx^n$ is greater than $0$, $\limsup_n\sqrt[n]{|a_n}<\infty$. Take $r>0$ such that $\limsup_n\sqrt[n]{|a_n|}<r$. Then, if $n\gg0$, $\sqrt[n]{|a_n|}<r$. Since the inequality $\sqrt[n]{|a_n|}\geqslant r$ can take place only for finitely many $n$'s, if you replace $r$ by some larger number $R$, you will have $(\forall n\in\Bbb N):\sqrt[n]{|a_n|}<R$. But then $|a_n|<R^n$, and so$$|b_n|=\left|\sum_{k=0}^na_k\right|\leqslant1+R+R^2+\cdots+R^n=\frac{R^{n+1}-1}{R-1}.$$But then$$\limsup_n\sqrt[n]{|b_n|}\leqslant R.$$

$\endgroup$
4
  • $\begingroup$ The geometric series would only converge if $|R|<1$ but you haven't proven this $\endgroup$
    – Jake
    Apr 7, 2021 at 23:05
  • 1
    $\begingroup$ Only a finite geometric sum is used in this proof. You don't need $|R|<1$. @Jake $\endgroup$ Apr 7, 2021 at 23:19
  • $\begingroup$ @KaviRamaMurthy Doesn't it become an infinite series in the last statement because you are taking the limit as n approaches infinity? $\endgroup$
    – Jake
    Apr 7, 2021 at 23:20
  • $\begingroup$ Never mind I understand what he meant $\endgroup$
    – Jake
    Apr 8, 2021 at 1:16
2
$\begingroup$

Here is a simple solution:

Notice that $$\Big(\sum_{n\geq0}a_nz^n\Big)\Big(\sum_{n\geq0}z^n\Big)=\sum_{n\geq0}b_nz^n$$

The series in the RHS converges in the ball $B(0;r)=\{z:|z|<r\}$ where $r=\min\{R_a,1\}$ and so, $r\leq R_b$. By assumption $R_a>0$, thus $R_b>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.