# Comparison Test for Convergence of Power Series

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## Theorem

Let $A = \displaystyle \sum_{n \mathop \ge 0} a_n z^n$ and $B = \displaystyle \sum_{n \mathop \ge 0} b_n z^n$ be power series in $\C$.

Let $R_A$ and $R_B$ be the radii of convergence of $A$ and $B$ respectively.

Let $\cmod {b_n} \le \cmod {a_n}$ for all $n \in \N$.

Then $R_A \le R_B$.

## Proof

Aiming for a contradiction, suppose $R_A > R_B$.

Let $z_0 \in \C$ such that $R_B < \cmod {z_0} < R_A$.

Then $A$ is convergent at $z_0$ but $B$ is divergent at $z_0$.

But by the Comparison Test, if $A$ is convergent at $z_0$ then $B$ is also convergent at $z_0$.

From this contradiction it follows that there can be no such $z_0$.

That is:

- $R_A \le R_B$

$\blacksquare$

## Sources

- 1960: Walter Ledermann:
*Complex Numbers*... (previous) ... (next): $\S 4$. Elementary Functions of a Complex Variable: Exercise $4$