Why the convergency of Cauchy product requires at least one of the series to be abs. convergent while the product of two convergent seq. converges?

Question

In baby rudin, Theorem 3.3(c) states,

suppose $\{s_n\}$,$\{t_n\}$ are complex sequences and $\lim_{n\rightarrow \infty}s_n=s$, $\lim_{n\rightarrow \infty}t_n=t$, then $\lim_{n\rightarrow \infty}s_nt_n=st$

here we don't require $\{s_{n+1}-s_n\}$ or $\{t_{n+1}-t_n\}$ to be absolutely convergent. But then Theorem 3.50 states,

if

• $\sum_{n=0}^\infty a_n$ converges absolutely,
• $\sum_{n=0}^{\infty}a_n=A$ and $\sum_{n=0}^{\infty}b_n=B$,
• $c_n=\sum_{k=0}^{n}a_kb_{n-k}$,

then $\sum_{n=0}^\infty c_n=AB$

which means $\lim_{n\rightarrow \infty}A_nB_n=AB$, where $A_n=\sum_{k=0}^na_k$ and $B_n=\sum_{k=0}^nb_k$.

But if we consider $\{A_n\}$ and $\{B_n\}$ as two sequences, it doesn't require $\sum_{n=0}^\infty a_n$ to be absolutely convergent to make sure $\lim_{n\rightarrow \infty}A_nB_n=AB$ according to Theorem 3.3(c). I'm quite confused about this. Any answer or hint would be greatly appreciated!

Answer 1

Yes, Theorem 3.3(c) proves that $$\lim_{n\to\infty} \left(\sum_{k=0}^n a_k\right)\left(\sum_{k=0}^n b_k\right) = AB$$ but that's not the Cauchy product:

$$\left(\sum_{k=0}^n a_k\right)\left(\sum_{k=0}^n b_k\right) \ne \sum_{k=0}^n\sum_{i=0}^k a_i b_{k-i}$$