Created on July 10, 2023

Written by Some author

Read time: 2 minutes

Summary: This passage presents the proof of a theorem stating that if two infinite series converge, their respective terms can be multiplied to obtain a new convergent series, even without the condition of absolute convergence.

Arithmetic operation over series.

Series are operating additionally and scaling nicely. However, the case for multiplication can be tricky. Let's look at the following theorem.

$\textbf{Theorem}$ If $\sum_{n=0}^{\infty} a_n$ converges absolutely, $\sum_{n=0}^{\infty} a_n=A$, $\sum_{n=0}^{\infty} b_n=B$, $c_n=\sum_{k=0}^n a_k b_{n-k} \quad(n=0,1,2, \ldots)$.

Then $ \sum_{n=0}^{\infty} c_n=A B$

Proof. let $A_n = \sum_{i=0}^n a_i$, $B_n = \sum_{i=0}^n b_i$, $C_n = \sum_{i=0}^n c_n$. Now we want to show that $ \sum_{n=0}^{\infty} c_n=A B$, which can be derived from $\lim_{n\to \infty}C_n = AB$,

which can be derived from $\lim_{n\to \infty}C_n =\lim_{n \to \infty}A_n B$, which can be derived from $\lim_{n\to \infty}C_n - A_n B = 0 $

Since we have $C_n -A_n B = \sum_{i=0}^n \sum_{k=0}^i a_k b_{i-k} -A_n B = \sum_{k=0}^n a_k \sum_{i=0}^{n-k}b_i-A_nB = \sum_{k=0}^n a_k \sum_{i=0}^{n-k}b_i-A_nB = \sum_{k=0}^n a_k B_{n-k}-A_nB$

$=\sum_{k=0}^n a_k (B - \sum_{i={n-k+1}}^\infty b_{i})-A_nB$

Let's denote $f(x) = -\sum_{i=x+1}^{\infty} b_i,$ then. we will have

$C_n -A_n B = \sum_{k=0}^n a_k (B +f(n-k))-A_nB = \sum_{k=0}^n a_k f(n-k)$

So all we want to show is as $n$ goes to infinity the right hand side $\sum_{k=0}^n a_k f(n-k)$ goes to zero.

We can accomplish this by noting as $x$ goes to infinity $f(x)$ goes to zero due to the fact that $B_n$ goes to $B$ as $n$ goes to infinity.

Then we will have for every $\epsilon >0$, there exists an $N$ such that for every $n' >N$, we will have $|f(n')|<\epsilon.$ Then, for $n >N$,

$|\sum_{k=0}^n a_k f(n-k)| = |\sum_{k=0}^n a_{n-k} f(k)| = |\sum_{k=0}^N a_{n-k} f(k)+ \sum_{k=N+1}^n a_{n-k} f(k)| \le |\sum_{k=0}^N a_{n-k} f(k)| +| \sum_{k=N+1}^n a_{n-k} f(k)| < |\sum_{k=0}^N a_{n-k} f(k)| + A' \epsilon$,

where $A'$ is the sum of $|a_n|$ and exists due to $\sum a_n$ converges absolutely.

Then if we sending $n$ to $\infty$, we will have

$$\limsup_{n \to \infty} \left|\sum_{k=0}^n a_k f(n-k)\right| \le A' \epsilon $$

due to $\epsilon$ being arbitrary, we have the limit to be zero. And therefore, we have the original.