# How to calculate the radius of convergence of this power series?

Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ be a formal pwer series with radius of convergence $R(f)=1$. Set $s_n=a_0+a_1+....a_n$. Let $g(z)=\sum_{n=0}^{\infty}s_nz^n$. Show that $R(g)=1$.

I noticed that $g(z)-zg(z)=f(z)$, if we are allowed to rearrange the summation of those power series $g(z)$ and $zg(z)$. But we don't know the radius of convergence yet so we can't say so. We can at most say that within $|z|<\min(1,R(g))$, the equation holds. So it means that radius of convergence of $g$ is greater or equal to $1$, but if $|z|>1$, the RHS $f(z)$ diverges, so the radius of convergence can only be $1$? I forgot to mention that the coefficients and $z$ take on $\mathbb{C}$

• Maybe it is helpful to refer to the definition of $R(g)$. Can you prove that if $R(g) < 1$, then the series converges and that if $R(g) > 1$, then the series diverges?
– user258700
Commented Nov 2, 2015 at 20:51
• Yeah...I edited my question, is it right now?
– jack
Commented Nov 2, 2015 at 20:54
• If you talk about convergence, then your series is not formal, so you should probably eliminate this word. You should also specify whether the coefficients are real or complex. Commented Nov 2, 2015 at 22:22

You are almost there. You want to deduce $(1-z)g(z) = f(z)$, which can be written $$g(z) = \frac{1}{1-z} f(z)$$ if $z \not= 1$. Recongnize the first term? The series defining $g$ is just the Cauchy product $$\left( \sum_{k=0}^\infty z^k \right) \left( \sum_{k=0}^\infty a_k z^k \right)$$ and both series in the product have radius of convergence $1$.
We can show a little more: For $r \in [0,1),$
$$\tag 1 \sum_{n=0}^{\infty}(|a_0|+ \cdots |a_n|)r^n < \infty.$$
Proof: Let $h(r) = \sum_{n=0}^{\infty} |a_n|r^n.$ Then $h(r)< \infty, r\in [0,1).$ Now fix $r \in [0,1).$ We have
$$\sum_{n=0}^{\infty}(|a_0|+ \cdots |a_n|)r^n = \sum_{n=0}^{\infty}(|a_0|+ \cdots + |a_n|)r^{n/2}\cdot r^{n/2}$$ $$\le \sum_{n=0}^{\infty}(|a_0|+ |a_1|r^{1/2} + |a_2|(r^{1/2})^2 + \cdots+ |a_n|(r^{1/2})^n)\cdot r^{n/2}$$ $$\le \sum_{n=0}^{\infty}h(r^{1/2})\cdot r^{n/2} = h(r^{1/2})/(1-r^{1/n})<\infty.$$