0
$\begingroup$

Is it sufficient to prove $P(x) \ge (\text{or} \le) a$ if we already know $P(x) > (\text{or} <)a$?

For example, to prove $$ \forall n \ge 1 , \sum_{i=1}^{n}\frac{1}{i^2} \le 2 $$

Suppose I have already proved $$ \forall n \ge 1 , \sum_{i=1}^{n}\frac{1}{i^2} < \frac{7}{4} - \frac{1}{n} $$ Then, are we done? (Because $\frac{7}{4}-\frac{1}{n} < 2 \le 2$)

$\endgroup$
1
$\begingroup$

Of course. If "greater than" is true, then "greater than or equal to" is true. Look at the truth table under OR here: https://medium.com/i-math/intro-to-truth-tables-boolean-algebra-73b331dd9b94.

$\endgroup$
  • $\begingroup$ Opps! I just found that $\le$ means less that or equal to in English, so it is actually kind of a logic union. $\endgroup$ – 王文军 or Wenjun Wang Jan 12 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.