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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?

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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$$\limsup_n\sqrt[n]{|b_n|}\leqslant R.$$

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  • $\begingroup$ The geometric series would only converge if $|R|<1$ but you haven't proven this $\endgroup$ – Jake 2 days ago
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    $\begingroup$ Only a finite geometric sum is used in this proof. You don't need $|R|<1$. @Jake $\endgroup$ – Kavi Rama Murthy 2 days ago
  • $\begingroup$ @KaviRamaMurthy Doesn't it become an infinite series in the last statement because you are taking the limit as n approaches infinity? $\endgroup$ – Jake 2 days ago
  • $\begingroup$ Never mind I understand what he meant $\endgroup$ – Jake 2 days ago
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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 $R_b>0$.

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