How to prove that 1 / n is O (1)?

Question

How to mathematically prove that 1 / n is O (1)? I am confused where to start. Any help?

Answer 1

As stated in this task, start by defining the Big-O notation.

F(x) = O(G(x)) IFF there exist constants k and m, 
such that for all n > m, k*|G(n)| > F(n). 

(Refer to the text tag for the exact wording here.)

Informally, this means that if we go far enough, then G (n) will dominate F (n), regardless of how big an initial advantage we give F (n) using constant factors.

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So how do you confirm something like this?

Evidence like this is usually carried out constructively - showing that certain well-chosen values ​​for m and k do the work of inequality.

. m k, . / , 1/n ( ), , m k .

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( Loadmaster): . 1/n = O (1) - , , "". , (, P, NP, EXP), .

Answer 2

, , (N >= 1), 1/2 :)

, 1/n n > 1, , , , n 1 !

Answer 3

, . f (n) = 1/n g (n) = 1. C k , 0 <= f (n) <= C * g (n), n > k. , C = 2 k = 0. n > 0 0 <= 1/n <= 2 * 1. C k 1/n O (1).

Is it necessary to prove this a little more rigorously? Well ... clear 1 / n <1 if and only if 1 <n (by simple algebra). Therefore, if n = 1 (which means 1 / n = 1), we know that 2 * 1> 1 and if n> 1, then this means that it means that 1 / n <1, in this case 2 * 1> 1 / n.

Answer 4

You really do not need to prove it, it is a fact. upper bound 1 / n, if n is a positive integer greater than 0, always 1.