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 $\frac{7}{4}-\frac{1}{n} < 2 \le 2$)


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.

  • $\begingroup$ 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

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