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?

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

  • 1
    $\begingroup$ @Metric, I am not sure if I am missing something, but the below need not always be true, $$M |\sum_{n=p}^{q-1} (b_n - b_{n+1}) + b_q + b_p| = M (\sum_{n=p}^{q-1} (b_n - b_{n+1}) + b_q + b_p),$$ $\endgroup$
    – texmex
    Commented Feb 13, 2020 at 9:48
  • $\begingroup$ Is convergence of $b_n$ to $0$ really required there? $\endgroup$
    – Koro
    Commented Mar 3, 2021 at 2:18

1 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.

  • $\begingroup$ Thank you very much for your answer. I wanna write $|r_1 c_1 + r_2 c_2| \leq M (c_1 + c_2)$ in (1) too, because $c_1 + c_2 \geq 0$. $\endgroup$
    – tchappy ha
    Commented Jan 26, 2019 at 2:36
  • 1
    $\begingroup$ I asked "Why did Rudin write $M |\sum_{n=p}^{q-1} (b_n - b_{n+1}) + b_q + b_p|$?". And I wanna ask you the same question. Why did you write $M |c_1 + c_2|$ instead of $M (c_1 + c_2)$ in (1)? $\endgroup$
    – tchappy ha
    Commented Jan 26, 2019 at 2:37
  • $\begingroup$ But I understand my question is about one's taste. Thank you. $\endgroup$
    – tchappy ha
    Commented Jan 26, 2019 at 2:40
  • $\begingroup$ It was really a result of two things: (1) taste, as you've mentioned, and (2) the more general fact where the $c_i$'s are real numbers not necessarily non-negative. In this case, the absolute value signs make the result more elegant and are needed. If you want me to show you what I mean more precisely and formally, then let me know! I'll edit the post appropriately. $\endgroup$
    – Metric
    Commented Jan 27, 2019 at 0:36
  • 3
    $\begingroup$ I'm not sure why this result applies to that above since $b_q-b_p$ is negative and does not imply necessarily that $b_q+b_p$ is positive. In particular if $0 \geq b_p \geq b_q$ their sum is not positive. $\endgroup$ Commented Jul 19, 2021 at 20:00

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