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Proving the Unboundedness of the Sequence $x_n frac{2^n}{n}$
Proving the Unboundedness of the Sequence $x_n frac{2^n}{n}$
In this article, we will delve into the proof that the sequence ( x_n frac{2^n}{n} ) is unbounded. To be precise, we will demonstrate that for any real number ( M ), there exists an integer ( n ) such that ( x_n > M ).
Steps to Prove Unboundedness
The initial sequence we will analyze is: [ x_n frac{2^n}{n} ] Our goal is to show that this sequence is not bounded, meaning for any given ( M in mathbb{R} ), there is an integer ( n ) such that ( x_n > M ).
Examining the Growth
As ( n ) increases, the term ( 2^n ) grows exponentially whereas ( n ) grows linearly. Intuitively, this suggests that ( x_n ) should increase without bound. To formalize this, we need to establish a more rigorous proof.
Formal Argument
Let us take any positive real number ( M ). We want to find an integer ( n ) such that: [ frac{2^n}{n} > M ] Rewriting this inequality, we get: [ 2^n > M cdot n ] This inequality suggests that the exponential growth of ( 2^n ) surpasses the linear growth of ( M cdot n ).
Using Limits
To provide a rigorous proof, we can use the concept of limits. Let us consider the limit of ( frac{2^n}{n} ) as ( n ) approaches infinity: [ lim_{n to infty} frac{2^n}{n} ] Applying L'H?pital's Rule, which is suitable because both the numerator and the denominator approach infinity, we get: [ lim_{n to infty} frac{2^n}{n} lim_{n to infty} frac{2^n ln 2}{1} infty ] This limit confirms that as ( n ) increases, the value of ( frac{2^n}{n} ) can be made arbitrarily large.
Conclusion
Since the limit of ( frac{2^n}{n} ) as ( n ) approaches infinity is infinity, this proves that the sequence ( x_n frac{2^n}{n} ) is unbounded. For any given ( M ), we can find an integer ( n ) such that ( x_n > M ).
Another Proof Strategy
Let ( R in mathbb{R} ). If ( n geq 2 ), then we have:
[ 2^n 1 - (-1)^n - binom{n}{2} R^n ]This implies:
[ x_n > -R ]Hence, the sequence ( langle x_n rangle ) is unbounded.
Proof by Contradiction
A sequence ( {x_n} ) of real numbers is bounded if there exists an ( M in mathbb{R} ) such that for all ( n in mathbb{N} ): [ x_n leq M ] If the sequence is not bounded, it is unbounded. So, our goal is to show that ( x_n frac{2^n}{n} ) is not bounded.
A proof by contradiction is a natural choice here. Suppose the sequence is bounded. Then, there exists an ( M in mathbb{R} ) such that for all ( n in mathbb{N} ): [ x_n leq M ] Since ( x_n > 0 ) for all ( n ), we can more simply write: [ x_n leq M ] This implies that for all ( n ): [ frac{2^n}{n} leq M ] Rearranging, we get: [ 2^n leq nM ]
Using the Binomial Theorem
The first step is to apply the Binomial Theorem: [ 2^n (1 1)^n sum_{k0}^n binom{n}{k} ] Keeping only the ( k2 ) term, we get: [ 2^n geq frac{n(n-1)}{2} ] By eliminating positive terms, we have made the left-hand side at least as large as the right-hand side. This shows that the exponential function is larger than a quadratic function as long as ( n geq 2 ).
Final Steps
Now, we need to show that for some ( n in mathbb{N} ), the quadratic term ( frac{n(n-1)}{2} ) gets larger than ( nM ): [ frac{n(n-1)}{2} geq nM ] Dividing both sides by ( n ), we get: [ frac{n-1}{2} geq M ] Since the natural numbers are unbounded, we can choose ( n ) such that: [ n geq 2M 1 ] Thus, the quadratic term is eventually larger than the linear term, leading to a contradiction. Therefore, the sequence ( x_n frac{2^n}{n} ) is unbounded.
Conclusion
By proving that the limit of ( frac{2^n}{n} ) as ( n ) approaches infinity is infinity, we have shown that the sequence ( x_n frac{2^n}{n} ) is unbounded. Any real number ( M ) cannot be a bound for this sequence, confirming its unbounded nature.