Prove that if $\sum a_n$, $\sum b_n$ and $\sum c_n$ converge to A,B and C then $C=AB$

Question

Define $\displaystyle c_n=\sum_{k=1}^n a_kb_{n+1-k}$. Prove that if $\sum a_n=A$, $\sum b_n=B$ and $\sum c_n=C$ then $C=AB$.

My teacher did not teach us Merten's Theorem or Abel's theorem. Is there a way to prove this without those theorems?

I don't know if this is the right approach, but I tried multiplying $\sum a_n$ and $\sum b_n$ and got

$$\sum a_n \sum b_n=a_0b_0+(a_0b_1+b_0a_1)+...+(a_0b_n+a_1b_{n-1}+...+a_{n-1}b_1+a_nb_0)$$

I think this is ultimately my goal: $$(\sum_{n=1}^N a_n)(\sum_{n=1}^N b_n)=\sum_{n=1}^N\sum_{k=1}^Na_n\cdot b_k=\sum_{n=1}^N\sum_{k=1}^na_k\cdot b_{n-k+1}$$

but I dont know how to show that $$\sum_{n=1}^N\sum_{k=1}^Na_n\cdot b_k=\sum_{n=1}^N\sum_{k=1}^na_k\cdot b_{n-k+1}$$

The terms that you calculated are precisely the $c_n$s (Cauchy product of $\sum_n a_n$ and $\sum_n b_n$ ). — Bernard, Dec 11 '20 at 01:29
I was told that $\sum c_n$ being convergent was enough to prove that statement and that $\sum a_n$ and $\sum b_n$ do not have to be absolutely convergent. — James Anderson, Dec 11 '20 at 01:36

score 2 · Answer 1 · answered Dec 11 '20 at 01:44

2

You can show that $\sum_n c_n$ is Cesaro-summable to $AB$, provided that $\sum_n a_n = A$, and $\sum_n b_n = B$ (Cesàro's theorem). If $\sum_n c_n$ is also summable, it must have sum $AB$.

answered Dec 11 '20 at 01:44

orangeskid

53,909

Prove that if $\sum a_n$, $\sum b_n$ and $\sum c_n$ converge to A,B and C then $C=AB$

1 Answers1