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.
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.
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)
thanks @MathMate
To prove that the sequence {-8 + (n / (n + 1))} is bounded, we need to show that there exists a real number M such that |Xn| ≤ M for all n ∈ N.
First, let's calculate the value of Xn to understand the pattern of the sequence.
Xn = -8 + (n / (n + 1))
To analyze the behavior of the sequence, we can consider the limit of Xn as n approaches infinity.
Taking the limit as n approaches infinity:
lim(n→∞) Xn = lim(n→∞) (-8 + (n / (n + 1)))
Since the numerator has only n and the denominator has n + 1, both tend to infinity as n approaches infinity. Therefore, we can use L'Hôpital's Rule to simplify the limit calculation.
Applying L'Hôpital's Rule:
lim(n→∞) (1 / 1) = 1
So, the limit of Xn as n approaches infinity is 1.
Now, we can choose M = 10 to prove that the sequence {-8 + (n / (n + 1))} is bounded.
Let's show that for all n ∈ N, |Xn| ≤ 10.
|Xn| = |-8 + (n / (n + 1)))|
Since -8 and (n / (n + 1)) are both negative for all n ∈ N, we can remove the absolute value sign:
|Xn| = -8 + (n / (n + 1))
Now, we need to show that -8 + (n / (n + 1)) ≤ 10 for all n ∈ N.
-8 + (n / (n + 1)) ≤ 10
To simplify this inequality, we can multiply both sides by (n + 1):
(n + 1)(-8 + (n / (n + 1))) ≤ (n + 1)(10)
Simplifying further:
-8(n + 1) + n ≤ 10(n + 1)
Expanding:
-8n - 8 + n ≤ 10n + 10
Combining like terms:
-7n - 8 ≤ 10n + 10
Adding 7n and 8 to both sides:
0 ≤ 17n + 18
Subtracting 18 from both sides:
-18 ≤ 17n
Dividing both sides by 17 (since 17 is positive):
-18/17 ≤ n
This shows that for all natural numbers n, n is greater than or equal to -18/17. In other words, the sequence {-8 + (n / (n + 1))} is bounded below by -18/17.
Since we have shown that -8 + (n / (n + 1)) ≤ 10 for all n ∈ N and -8 + (n / (n + 1)) is also bounded below by -18/17, we can conclude that the sequence {-8 + (n / (n + 1))} is bounded.
Therefore, we have successfully proved that the sequence {-8 + (n / (n + 1))} is bounded.