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,


  • $\sum_{n=0}^\infty a_n$ 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!


1 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:

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


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