A simple question about the proof of theorem 3.42 in Walter Rudin's "Principles of Mathematical Analysis"

Question

I am reading "Principles of Mathematical Analysis" by Walter Rudin.

I read the proof of theorem 3.42 on p.71.

And I have a simple question about the following calculation:

$$|\sum_{n=p}^{q} a_n b_n| = |\sum_{n=p}^{q-1} A_n (b_n - b_{n+1}) + A_q b_q - A_{p-1} b_p| \leq M |\sum_{n=p}^{q-1} (b_n - b_{n+1}) + b_q + b_p|.$$

My calculation is here:

$$|\sum_{n=p}^{q} a_n b_n| = |\sum_{n=p}^{q-1} A_n (b_n - b_{n+1}) + A_q b_q - A_{p-1} b_p| \leq \\\sum_{n=p}^{q-1} |A_n| |(b_n - b_{n+1})| + |A_q| |b_q| + |A_{p-1} | |b_p| \leq \\\sum_{n=p}^{q-1} M |(b_n - b_{n+1})| + M |b_q| + M |b_p| = \\\sum_{n=p}^{q-1} M (b_n - b_{n+1}) + M b_q + M b_p = \\ M (\sum_{n=p}^{q-1} (b_n - b_{n+1}) + b_q + b_p).$$

Of course,

$$M |\sum_{n=p}^{q-1} (b_n - b_{n+1}) + b_q + b_p| = M (\sum_{n=p}^{q-1} (b_n - b_{n+1}) + b_q + b_p),$$

but why did Rudin write as follows?

$$M |\sum_{n=p}^{q-1} (b_n - b_{n+1}) + b_q + b_p|.$$

Answer 1

You're calculation is fine.

Rudin used a fact which made his proof a bit shorter; it's an intuitive fact but I'll just prove it for completeness. I'll prove it for the case of two numbers for sake of readability and convenience. The argument for the general case is directly analogous.

Claim: Suppose $r_1, r_2 \in \mathbb{R}$ such that $|r_1|,|r_2| \le M$ for some $M \ge 0$. Suppose also that $c_1,c_2 \ge 0$. Then, $$ \lvert r_1c_1 + r_2c_2 \rvert \le M\lvert c_1 + c_2 \rvert \tag 1$$

Proof: Since $c_i \ge 0$, we get $$ -Mc_1 \le r_1c_1 \le Mc_1$$ $$ -Mc_2 \le r_2c_2 \le Mc_2$$ which implies $$-M(c_1 + c_2) \le r_1c_1 + r_2c_2 \le M(c_1 + c_2)$$ and since $|M| = M$, we get $(1)$.

Note that using the above fact makes the proof shorter than using the Triangle Inequality, but again, you're proof is still fine.