Question

definiton: A set S subsect of real number,
sequence {Xn} is called bounded if there is a number M such that |X|≤ M for all x ∈ S

prove the sequence (-8+(n/(n+1))) is bounded.

my anwser
proof: Let Xn = (-8+(n/(n+1)))
there exist M in real number such that |Xn| ≤ M for all n ∈ N.

Answer 1

Try and improve on the following.
R=set of all real numbers.
N=set of all natural numbers

Let Xn=-8+n/(n+1), where X∈R
then
Xn=-8 + (n+1)/(n+1) - 1/(n+1)
=-7 - 1/(n+1)
=>
|Xn|≤|-7|+|1/(n+1)| (triangle inequality)

But |1/(n+1)|≤1 ∀n∈N, therefore
|Xn|≤|-7|+|1|
=>
|Xn|≤8

Hence, we have just demonstrated that for M=8, |X|≤M ∀x∈R, which implies that {Xn} is bounded.

You may need to modify the proof due to the fact that R was used instead of S, where it was given that S is a subset of R.

Answer 2

Correction:
Somewhere the -8 was inadvertently written as -7, so after corrections,
|Xn|≤|-8|+|-1/(n+1)| ∀n∈N
or
|Xn|≤9 (or M=9)

Answer 3

thanks @MathMate