On the proof of Thm.3.50 in Baby Rudin

I have a question on the proof of Thm.3.50 in Rudin's Principles of Mathematical Analysis. This theorem is about the convergence of the Cauchy product of two infinite series. For convenience, I write the necessary details of the theorem and proof.

Theorem

Suppose

(a) $$\sum_{n = 0}^{\infty}a_n$$ converges absolutely,

(b) $$\sum_{n = 0}^{\infty}a_n = A$$,

(c) $$\sum_{n = 0}^{\infty}b_n = B$$

(d) $$c_n = \sum_{k = 0}^{n}a_kb_{n - k}.$$

Then, \begin{align*} \sum_{n = 0}^{\infty}c_n = AB. \end{align*}

Proof

Put \begin{align*} A_n = \sum_{k = 0}^{n}a_k, \ \ B_n = \sum_{k = 0}^{n}b_k, \ \ C_n = \sum_{k = 0}^{n}c_k, \ \ \beta_n = B_n - B. \end{align*} Then we can rewtrite $$C_n$$ as \begin{align*} C_n = A_nB + a_0\beta_n + a_1\beta_{n-1} + \cdots + a_n\beta_0. \end{align*} Put \begin{align*} \gamma_n = a_0\beta_n + a_1\beta_{n-1} + \cdots + a_n\beta_0 \\ \end{align*} so that \begin{align*} C_n = A_nB + \gamma_n. \end{align*} Our aim here is to show $$C_n \to AB$$. Since $$A_nB \to AB$$, it suffices to show \begin{align*} \gamma_n \to 0. \end{align*} Put \begin{align*} \alpha = \sum_{n = 0}^{\infty}|a_n|. \end{align*} By (c), we have that $$\beta_n \to 0$$, or that for any $$\varepsilon > 0$$, there is some integer $$N$$ such that $$|\beta_n| \leq \varepsilon$$ for $$n \geq N$$. Hence, for sufficiently large n \begin{align} |\gamma_n| &\leq |\beta_0a_n + \cdots + \beta_Na_{n - N}| + |\beta_{N + 1}a_{n - N - 1} + \cdots + \beta_na_0| \\ &\leq |\beta_0a_n + \cdots + \beta_Na_{n - N}| + \varepsilon\alpha. \tag{1} \end{align} Keeping $$N$$ fixed, and letting $$n \to \infty$$, we get \begin{align} \limsup_{n \to \infty} |\gamma_n| \leq \varepsilon\alpha, \tag{2} \end{align} since $$a_n \to 0$$ as $$n \to \infty$$. Since $$\varepsilon$$ is arbitrary, we obtain $$\lim_{n \to \infty}\gamma_n = 0.$$ (end)

Question

The quetion I have here is about (2). I see that, by triangular inequality and the fact that $$|a_n| \to 0$$, the term $$|\beta_0a_n + \cdots + \beta_Na_{n - N}|$$ goes to zero with $$N$$ fixed. Because we take the limit of (1), it seems to me that we obtain \begin{align} \lim_{n \to \infty} |\gamma_n| \leq \varepsilon\alpha \tag{3} \end{align} instead of (2). I notice that we get (3) if (2) holds since $$|\gamma_n|$$ is nonnegative, but I feel we can derive (3) directly from taking the limit of (1). Am I wrong in some points?

Any comment would be appriciated. Thank you!

If $$|t_n| \leq s_n$$ and $$\lim s_n=s$$ we cannot say that $$\lim t_n$$ exists. What we can say is $$\lim \sup |t_n| \leq \lim \sup s_n=\lim s_n=s$$ since $$\lim \sup$$ and $$\lim$$ are the same for a convergent sequence $$(s_n)$$.
Example: Let $$t_n=0$$ for $$n$$ even and $$t_n=1$$ for $$n$$ odd. Let $$s_n=1$$ for all $$n$$. Then $$\lim t_n$$ does not exist even thoug $$0 \leq t_n \leq s_n$$ and $$s_n \to 1$$.