Proving the Inequality of the Sequence ( left(1 - frac{1}{n}right)^n leq 3 ) for All ( n in mathbb{N} )

When dealing with mathematical sequences and inequalities, it is often necessary to rigorously prove certain properties. In this detailed guide, we will explore how to prove that the sequence ( s_n left(1 - frac{1}{n}right)^n ) is bounded by 3 for all natural numbers ( n ). This involves a combination of mathematical techniques, including the binomial theorem and series estimation.

Understanding the Context

The sequence in question is ( s_n left(1 - frac{1}{n}right)^n ). It is well-known that as ( n ) approaches infinity, this sequence converges to the number ( e ) (Euler's number) where ( e approx 2.71828 ). However, for finite values of ( n ), the sequence is bounded by 3. Our goal is to prove this statement rigorously.

Proving the Inequality

To prove that ( left(1 - frac{1}{n}right)^n leq 3 ) for all ( n in mathbb{N} ), we will use a few key steps involving estimation and the binomial theorem.

Step 1: Using the Binomial Theorem

The first step is to expand the expression using the binomial theorem.

Recall the binomial formula:

[ sum_{k0}^n {n choose k} left(-frac{1}{n}right)^k 1 - frac{n-1}{n} frac{(n-1)(n-2)}{2n^2} - cdots (-1)^n frac{1}{n^n} ]

For easier manipulation, we can write:

[ s_n 1 - frac{n-1}{n} frac{(n-1)(n-2)}{2n^2} - cdots (-1)^n frac{1}{n^n} ]

Now, we need to estimate the upper bound of this series.

Step 2: Estimating the Series

Let's estimate the terms of the series. Consider the general term:

[ {n choose k} left(-frac{1}{n}right)^k frac{n!}{k!(n-k)!} left(-frac{1}{n}right)^k ]

For simplicity, let's focus on the positive terms:

[ {n choose k} frac{1}{n^k} leq frac{1}{2^k} ]

This is because the binomial coefficient can be bounded by powers of 2 as ( n ) becomes large. This bounding allows us to simplify the series:

[ 1 - frac{1}{2} frac{1}{4} - frac{1}{8} cdots leq 1 frac{1}{2} frac{1}{4} cdots ]

The right-hand side is a geometric series with the sum:

[ frac{1}{1 - frac{1}{2}} 2 ]

Therefore, the sequence ( s_n ) is bounded above by 3 for all ( n ).

Additional Approaches to Prove the Inequality

Another way to prove the inequality is by showing that the sequence is monotonically increasing and finding an upper bound for the initial terms.

Show Monotonicity and Initial Terms

We claim that the sequence is monotonically increasing from a certain ( n ). To show this, consider:

[ s_{n 1} left(1 - frac{1}{n 1}right)^{n 1} ]

Using the binomial theorem, we can expand this and show that it is greater than or equal to ( s_n ) for ( n geq 3 ).

Additionally, we can find the upper bound for the initial terms. By bounding the terms appropriately, we can ensure that the sequence is always less than 3.

Conclusion

In conclusion, the sequence ( s_n left(1 - frac{1}{n}right)^n ) is bounded by 3 for all natural numbers ( n ). This can be proved rigorously using the binomial theorem and series estimation techniques. The sequence converges to ( e ) as ( n ) approaches infinity, but for finite ( n ), the bound remains at 3.

To summarize, the proof involves expanding the sequence using the binomial theorem, estimating the terms, and showing that the sequence is monotonically increasing and bounded properly.