Consider a complex power series $$s(z) = \sum_{n\geq 0} a_nz^n$$ with radius of covergence of $\rho(s)=1$. Then, consider the following power series: $$t(z) = \sum_{n\geq 0} b_nz^n $$ where $b_n = a_0 + a_1 + ....+a_n$. Prove that: $\rho(t)=1$.

My approach:

Consider the partial sum of the absolute values: $$ P_k = \sum_{n\geq 0}^{k} |b_0| + |b_1z| + ...+|b_kz^k| = |a_0| + |(a_0+a_1)z|+...+|(a_0 + ... + a_k)|z^k $$ It follows: $$ P_k \leq |a_0| + |a_0z| + |a_1z| + ... + |a_0z^k| +...+|a_kz^k| = |a_0|(1 + |z| +...+|z^k|) + |a_1z|(1+...+|z^{k-1}|)+....|a_kz^k| \leq |a_0|(1 + |z| +...+|z^k|) + |a_1z|(1+...+|z^{k}|)+....|a_kz^k|(1+|z|+...|z^k|) = (|a_0|+|a_1z|+...+|a_kz^k|)(1+|z|+...|z^k|) $$ From that inequality I get $\rho(t)\geq 1$. My question is: how to continue? Is my thought process and conclusion ok? since I'm working with the absolute values power series and not the actual power series. Thanks


Your argument in the one direction looks fine. For the other direction, we want to show

$$\limsup |a_0 + \cdots + a_n|^{1/n} \ge \limsup |a_n|^{1/n} .$$

Choose a subsequence $n_k$ such that $|a_{n_k}|^{1/{n_k}} \to \limsup |a_n|^{1/n}.$ Suppose $|a_0 + \cdots + a_{n_k-1}| \ge |a_{n_k}|/2 $ for infinitely many $k.$ For these $k$ we then have

$$|a_0 + \cdots + a_{n_k-1}|^{1/(n_k-1)} \ge (|a_{n_k}|/2)^{1/(n_k-1)}.$$

As $k\to \infty,$ the term on the right converges to $\limsup |a_n|^{1/n} $ and we're done.

Otherwise $|a_0 + \cdots + a_{n_k-1}| \le |a_{n_k}|/2 $ for infinitely many $k.$ In this case

$$|a_0 + \cdots + a_{n_k}|\ge |a_{n_k}|/2$$

for infinitely many $k.$ Raising both sides to the $1/n_k$ power then gives the result.


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