> [!NOTE] Theorem (The product of power series) > If [[Power Series|power series]] $\sum a_{n}x^{n}$ and $\sum b_{n} x^{n}$ converge for $x\in(-R,R)$, then so does the series $\sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} a_{k } b_{n-k} \right) x^{n} = \left( \sum_{i=0}^{\infty} a_{i } x^{i} \right) \left( \sum_{i=0}^{\infty} b_{j} x^{j} \right) $ **Proof**: Note that by [[Product of Sums]], $\left( \sum_{n=0}^{N} a_{n} x^{n} \right)\left( \sum_{m=0}^{N} b_{m}x^{m} \right)=\sum_{\substack{0\leq n \leq N \\ 0 \leq m \leq N}} a_{n}b_{m}x^{n+m}$ Consider $\begin{align} &\sum_{n=0}^{2N} \left( \sum_{m=0}^{n} a_{m}b_{n-m} \right)x^{n} - \left( \sum_{n=0}^{N} a_{ n}x^{n} \right) \left( \sum_{m=0}^{N} b_{m}x^{m} \right) \\ &= \sum_{n+m \leq 2N} a_{n} b_{m} x^{n+m} - \sum_{\substack{n\leq N \\ m \leq N}} a_{n}b_{m} x^{n+m} \\ &= \sum_{\substack{n \geq N+1\\ n+m \leq 2N}} a_{n} b_{m} x^{n+m} + \sum_{\substack{m \geq N+1\\ n+m \leq 2N}} a_{n} b_{m} x^{n+m} \\ &\leq \sum_{\substack{n \geq N+1\\ n+m \leq 2N}} |a_{n}| |b_{m}| |x|^{n+m} + \sum_{\substack{m \geq N+1\\ n+m \leq 2N}} |a_{n}| |b_{m}| |x|^{n+m} \\ &\leq \sum_{\substack{n \geq N+1\\ m \geq 0}} |a_{n}| |b_{m}| |x|^{n+m} + \sum_{\substack{m \geq N+1\\ n+m \geq 0}} |a_{n}| |b_{m}| |x|^{n+m} \\ &= \left( \sum_{n\geq N+1} |a_{n}| |x|^{n} \right) \left( \sum_{m \geq 0} |b_{m}| |x|^{m} \right) + \left( \sum_{m \geq N+1} |b_{m}||x|^{m} \right) \left( \sum_{n\geq 0} |a_{n}| |x|^{n} \right)\tag{*} \end{align}$ Now by [[Radius of Convergence of Absolute Real Power Series About Zero]], $\sum_{m\geq 0} |b_{m}|x^{m}$ and $\sum_{n\geq 0}|a_{n}|x^{n}$ are convergent. WLOG $\sum_{m \geq 0}|b_{m}||x|^{m}$ is the first series evaluated at $|x|.$ So $(*)$ converges to zero as $N\to \infty.$ ***Proof***: The result follows from directly from [[Merten's Convergence Theorem]] since by [[Radius of Convergence of Absolute Real Power Series About Zero]], both series **converge absolutely** in the interval of convergence.