What is the upper bound of the infinite sequence 1/x^2, where xE{R\0}
We know that x={1, 1/4, 1/9, 1/16,.....}
Since all x here are greater than 0 or less than equal to 1.
Can't we take 1 as an upper bound here(My tutorial instructor has mentioned so in an answer sheet) I'm confused. Can someone explain? Thanks!
Clearly 1 is the least upper bound, since all of the other terms are less than 1, and in fact the largest term is exactly 1.
Thank you very much!
To find the upper bound of the infinite sequence 1/x^2, where x ∈ {R\0}, we need to determine the largest value that the sequence can attain.
Let's examine the terms in the sequence: 1/x^2, where x ∈ {1, 1/4, 1/9, 1/16, ...}.
When x is 1, the term is 1/1^2 = 1/1 = 1.
When x is 1/4, the term is 1/(1/4)^2 = 1/(1/16) = 16.
When x is 1/9, the term is 1/(1/9)^2 = 1/(1/81) = 81.
When x is 1/16, the term is 1/(1/16)^2 = 1/(1/256) = 256.
As you can see, as x becomes smaller, the term becomes larger. The smallest x can get is 0, but it is excluded from the sequence. Therefore, the sequence becomes larger and larger as x approaches 0, but it never reaches or exceeds any finite number.
Therefore, we can conclude that there is no upper bound for the infinite sequence 1/x^2, where x ∈ {R\0}. The terms in the sequence can keep getting larger and larger indefinitely. In this case, your tutorial instructor may have meant that 1 can be used as an upper bound for certain values of x, but not for the entire sequence.