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 $s_n \to 1$.
