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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

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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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