# Convergence of power series with sum of coefficients

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?

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 $$\limsup_n\sqrt[n]{|a_n|}. Then, if $$n\gg0$$, $$\sqrt[n]{|a_n|}. 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 $$(\forall n\in\Bbb N):\sqrt[n]{|a_n|}. But then $$|a_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$$\limsup_n\sqrt[n]{|b_n|}\leqslant R.$$

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

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 $$B(0;r)=\{z:|z| where $$r=\min\{R_a,1\}$$ and so, $$r\leq R_b$$. By assumption $$R_a>0$$, thus $$R_b>0$$.