Convergent Sequence: Xn = (1+1/n)^n Explained

Hey everyone! Let's tackle a classic problem in real analysis: proving the convergence of the sequence xn = (1 + 1/n)^n. This sequence is super important because it creeps up in a lot of areas, and it’s closely linked to the definition of the number e. So, let’s break it down step-by-step and make sure we really understand what’s going on.

Understanding the Sequence

Before we dive into the nitty-gritty proof, let's take a moment to understand what this sequence actually represents. We're essentially taking the number 1, adding a fraction (1/n) to it, and then raising the whole thing to the power of n. As n gets bigger and bigger, the fraction 1/n gets smaller and smaller, approaching zero. So, we're adding a tiny amount to 1, but then raising it to a large power. It's like compound interest – a small increase applied repeatedly can lead to significant growth. Our goal is to show that this growth doesn't go on forever; instead, the sequence settles down and converges to a specific value, which, as you might have guessed, is e.

To really grasp this, imagine n representing the number of times interest is compounded in a year. If you start with $1 and get 100% interest per year, compounding annually (n = 1) gives you $2 at the end of the year. Compounding semi-annually (n = 2) gives you $(1 + 1/2)^2 = \$2.25$. As you compound more and more frequently, the return increases, but it doesn't increase indefinitely. The sequence (1 + 1/n)^n captures this behavior perfectly. This intuition is crucial because it guides our approach to proving convergence. We expect the sequence to increase, but we also expect it to be bounded above, which are the two key ingredients for proving convergence using the Monotone Convergence Theorem.

The Binomial Theorem Approach

Okay, so how do we actually prove that this sequence converges? A common starting point, and the one mentioned in the original question, is using the binomial theorem. This theorem gives us a way to expand expressions of the form (a + b)^n, and it's particularly helpful here because it allows us to break down the sequence into a sum of terms that we can analyze individually. The binomial theorem states:

(a + b)^n = Σ (n choose k) * a^(n-k) * b^k (where the summation is from k = 0 to n)

where (n choose k) is the binomial coefficient, also written as nCk or (n!)/(k!(n-k)!).

Applying this to our sequence, where a = 1 and b = 1/n, we get:

(1 + 1/n)^n = Σ (n choose k) * 1^(n-k) * (1/n)^k = Σ (n choose k) * (1/n)^k (summation from k = 0 to n)

Expanding the binomial coefficient, we have:

(n choose k) = n! / (k! * (n - k)!) = [n * (n - 1) * (n - 2) * ... * (n - k + 1)] / k!

So, the sequence becomes:

(1 + 1/n)^n = Σ [n * (n - 1) * (n - 2) * ... * (n - k + 1)] / (k! * n^k) (summation from k = 0 to n)

This looks a bit intimidating, but let’s simplify each term in the sum. We can rewrite the numerator as:

[n * (n - 1) * (n - 2) * ... * (n - k + 1)] / n^k = 1 * (1 - 1/n) * (1 - 2/n) * ... * (1 - (k - 1)/n)

Now, our sequence looks like this:

(1 + 1/n)^n = Σ [1 * (1 - 1/n) * (1 - 2/n) * ... * (1 - (k - 1)/n)] / k! (summation from k = 0 to n)

This is a crucial step! We've transformed the original expression into a sum of terms, each of which has a clear structure. Each term involves a product of factors that are all less than or equal to 1, divided by k!. This form is much easier to analyze than the original.

Proving Monotonicity

Now that we have this expanded form, we can tackle the first key part of proving convergence: showing that the sequence is monotonically increasing. In simpler terms, we want to show that each term in the sequence is greater than or equal to the previous term. Mathematically, this means we need to show that x_(n+1) ≥ x_n for all n.

Let's write out the expression for x_(n+1) using our expanded form:

x_(n+1) = Σ [1 * (1 - 1/(n+1)) * (1 - 2/(n+1)) * ... * (1 - (k - 1)/(n+1))] / k! (summation from k = 0 to n+1)

Comparing this to the expression for x_n, we see that each term in the sum for x_(n+1) looks very similar to the corresponding term in the sum for x_n. The key difference is that the factors (1 - j/n) in x_n are replaced by (1 - j/(n+1)) in x_(n+1), where j ranges from 1 to k - 1. Since n + 1 > n, we have j/(n+1) < j/n, and therefore (1 - j/(n+1)) > (1 - j/n). This means that each individual term in the sum for x_(n+1) is greater than the corresponding term in the sum for x_n.

Furthermore, the sum for x_(n+1) has one extra term (the term for k = n + 1) that is not present in the sum for x_n. This extra term is positive, so it further contributes to making x_(n+1) larger. Putting it all together, we can confidently conclude that x_(n+1) > x_n for all n. This confirms that the sequence is strictly increasing.

Proving Boundedness

The second crucial step in proving convergence is showing that the sequence is bounded above. This means we need to find a number M such that x_n ≤ M for all n. If a sequence is both increasing and bounded above, the Monotone Convergence Theorem guarantees that it converges.

Let's go back to our expanded form of the sequence:

x_n = Σ [1 * (1 - 1/n) * (1 - 2/n) * ... * (1 - (k - 1)/n)] / k! (summation from k = 0 to n)

We know that each factor (1 - j/n) is less than or equal to 1. Therefore, we can replace each of these factors by 1 without decreasing the value of the sum. This gives us the following inequality:

x_n ≤ Σ 1 / k! (summation from k = 0 to n)

Now, we have a much simpler sum to deal with: the sum of the reciprocals of the factorials. This sum is closely related to the Taylor series expansion of the exponential function, e^x. Recall that the Taylor series for e^x is:

e^x = Σ x^k / k! (summation from k = 0 to infinity)

Setting x = 1, we get:

e = Σ 1 / k! (summation from k = 0 to infinity)

Our sum for x_n is just a partial sum of this infinite series. Since all the terms in the series are positive, the partial sums are increasing and bounded above by the full sum, which is e. Therefore, we have:

x_n ≤ Σ 1 / k! ≤ e (summation from k = 0 to n)

This shows that the sequence x_n is bounded above by e. And since e is approximately 2.71828, we have a concrete upper bound for our sequence.

Conclusion: Convergence Achieved!

We've done it! We've shown that the sequence xn = (1 + 1/n)^n is both monotonically increasing and bounded above. By the Monotone Convergence Theorem, this means that the sequence must converge to a limit. We haven't explicitly calculated the limit here, but we know that it exists and is less than or equal to e. In fact, the limit is equal to e, which is a fundamental constant in mathematics.

So, to recap, we used the binomial theorem to expand the sequence, analyzed the resulting terms to prove monotonicity, and then bounded the sequence above using the Taylor series for e. This is a classic example of how we can combine different mathematical tools to solve a problem in real analysis. I hope this explanation has been helpful and has given you a deeper understanding of this fascinating sequence!