Limit Of Nth Root Of N Factorial Calculation And Solution

Hey math enthusiasts! Today, we're diving into a fascinating problem that combines limits, nth roots, and factorials. Buckle up, because we're going to unravel the mystery behind the limit of the nth root of n factorial and then use that knowledge to calculate a specific expression. This is going to be an exciting journey, so let's get started!

Understanding the Problem

Let's break down the problem statement. We're given that the limit as n approaches infinity of the nth root of the product of the first n natural numbers (which is simply n factorial) equals L. In mathematical notation:

lim (n→∞) ⁿ√(1 ⋅ 2 ⋅ 3 ⋅ ... ⋅ n) = L

This L represents the value that the nth root of n factorial approaches as n gets incredibly large. Our mission is to find this L. We're also told that the negative natural logarithm of L is equal to p:

-ln L = p

Finally, the ultimate goal is to determine the value of the expression (1/(2p))^2. So, it's a multi-step problem: first, we need to find L, then p, and finally, plug p into the expression to get our answer.

Diving Deep into Factorials and nth Roots

Before we jump into the solution, let's take a moment to appreciate the players involved: factorials and nth roots. The factorial function, denoted by n!, is the product of all positive integers less than or equal to n. For example, 5! = 5 * 4 * 3 * 2 * 1 = 120. Factorials grow incredibly fast, which makes them interesting to study in the context of limits.

The nth root, on the other hand, is the inverse operation of raising something to the power of n. The nth root of a number x is the number that, when raised to the power of n, equals x. For instance, the cube root (3rd root) of 8 is 2 because 2^3 = 8.

Combining these two concepts, the nth root of n factorial, creates a fascinating mathematical challenge. We're essentially trying to find a kind of 'average' growth rate of the factorial function as n gets larger and larger. How do we tackle this beast? Well, that's where some clever techniques come into play.

The Power of Stirling's Approximation

One of the most powerful tools in our arsenal for dealing with factorials in limits is Stirling's approximation. This approximation provides a close estimate for the value of n! when n is large. There are several forms of Stirling's approximation, but the one that's most useful for our problem is:

n! ≈ √(2πn) * (n/e)^n

where e is the base of the natural logarithm (approximately 2.71828). This approximation tells us that n! grows roughly proportionally to the square root of 2πn multiplied by (n/e) raised to the power of n. It's a remarkable result that connects factorials to continuous functions like the exponential function.

Why is Stirling's approximation so important? Because it allows us to replace the factorial function, which is discrete in nature (defined only for integers), with a continuous function that's much easier to work with in the context of limits. We can apply calculus techniques, such as L'Hôpital's rule, which are designed for continuous functions.

Solving for L

Now that we have Stirling's approximation, let's use it to find the value of L. We start by substituting the approximation into our limit:

L = lim (n→∞) ⁿ√(n!) ≈ lim (n→∞) ⁿ√(√(2πn) * (n/e)^n)

To simplify this, let's rewrite the nth root as a power of 1/n:

L ≈ lim (n→∞) (√(2πn) * (n/e)^n)^(1/n)

Using the properties of exponents, we can distribute the power of 1/n:

L ≈ lim (n→∞) (2πn)^(1/(2n)) * (n/e)^(n*(1/n))

Simplifying further:

L ≈ lim (n→∞) (2πn)^(1/(2n)) * (n/e)

Now, let's analyze the first term, (2πn)^(1/(2n)). As n approaches infinity, the exponent 1/(2n) approaches 0. Any positive number raised to the power of 0 approaches 1. So, we have:

lim (n→∞) (2πn)^(1/(2n)) = 1

This leaves us with:

L ≈ lim (n→∞) 1 * (n/e) = lim (n→∞) n/e

Oops! It seems like we've hit a snag. If we directly substitute infinity into n/e, we get infinity. This indicates there might be an issue with how we applied Stirling's approximation or with our initial approach. We need to be more careful with the limit. Let's take a step back and try a different tactic.

A Logarithmic Approach: A Clever Twist

Sometimes, when dealing with limits involving exponents and roots, a clever trick is to take the natural logarithm of both sides. This can often simplify the expression and make the limit easier to evaluate. So, let's do that! We'll take the natural logarithm of our limit expression:

ln L = ln [lim (n→∞) ⁿ√(n!)]

Since the natural logarithm is a continuous function, we can move it inside the limit:

ln L = lim (n→∞) ln [ⁿ√(n!)]

Now, let's rewrite the nth root as a power of 1/n and use the property of logarithms that ln(a^b) = b * ln(a):

ln L = lim (n→∞) ln [(n!)^(1/n)] = lim (n→∞) (1/n) * ln(n!)

This looks more manageable. We still have ln(n!), but at least it's not inside a root. Now, let's express ln(n!) as a sum of logarithms:

ln(n!) = ln(1 * 2 * 3 * ... * n) = ln(1) + ln(2) + ln(3) + ... + ln(n) = Σ[k=1 to n] ln(k)

Substituting this back into our limit expression:

ln L = lim (n→∞) (1/n) * Σ[k=1 to n] ln(k)

Aha! This expression looks very much like a Riemann sum! A Riemann sum is a way to approximate the definite integral of a function. Recall that the definite integral of a function f(x) from a to b can be approximated by the sum of the areas of rectangles under the curve:

∫[a to b] f(x) dx ≈ Σ[i=1 to n] f(x_i) * Δx

where x_i are points in the interval [a, b] and Δx is the width of each rectangle. In our case, we have (1/n) multiplied by a sum of ln(k) values. This strongly suggests that we can express our limit as a definite integral.

Connecting the Sum to an Integral

To make the connection to a Riemann sum more clear, let's rewrite our sum slightly:

ln L = lim (n→∞) (1/n) * Σ[k=1 to n] ln(k) = lim (n→∞) Σ[k=1 to n] ln(k) * (1/n)

Now, we can think of 1/n as Δx, the width of each rectangle. The terms ln(k) can be seen as the values of a function at certain points. Let's consider the function f(x) = ln(x). If we divide the interval [0, 1] into n equal subintervals, each of width 1/n, the right endpoints of these subintervals will be 1/n, 2/n, 3/n, ..., n/n = 1. So, we can rewrite the sum as:

ln L = lim (n→∞) Σ[k=1 to n] ln(k/n) * (1/n)

This is now a Riemann sum for the integral of ln(x) from 0 to 1! However, there's a slight complication: ln(0) is undefined. We need to shift our interval slightly to avoid this issue. Let's consider the integral from 1/n to 1 instead of 0 to 1. As n approaches infinity, 1/n approaches 0, so this shift won't affect the limit. Our Riemann sum now corresponds to the integral:

ln L = ∫[0 to 1] ln(x) dx

Evaluating the Integral

Now we have a definite integral to evaluate. The integral of ln(x) can be found using integration by parts. Let u = ln(x) and dv = dx. Then, du = (1/x) dx and v = x. Using the integration by parts formula, ∫u dv = uv - ∫v du:

∫ ln(x) dx = x ln(x) - ∫ x * (1/x) dx = x ln(x) - ∫ 1 dx = x ln(x) - x + C

where C is the constant of integration. Now we can evaluate the definite integral:

ln L = ∫[0 to 1] ln(x) dx = [x ln(x) - x] evaluated from 0 to 1

Let's plug in the limits of integration:

ln L = (1 * ln(1) - 1) - lim (a→0+) (a ln(a) - a)

We need to be careful with the limit as a approaches 0. We know that ln(1) = 0, so the first term is -1. For the limit, we have an indeterminate form of the type 0 * (-∞). To evaluate this, we can rewrite a ln(a) as ln(a)/(1/a) and apply L'Hôpital's rule:

lim (a→0+) a ln(a) = lim (a→0+) ln(a) / (1/a)

The derivative of ln(a) is 1/a, and the derivative of 1/a is -1/a^2. Applying L'Hôpital's rule:

lim (a→0+) ln(a) / (1/a) = lim (a→0+) (1/a) / (-1/a^2) = lim (a→0+) -a = 0

So, the limit of a ln(a) as a approaches 0 is 0. Also, the limit of a as a approaches 0 is 0. Therefore:

ln L = (0 - 1) - (0 - 0) = -1

Finding L, p, and the Final Answer

We've found that ln L = -1. To find L, we take the exponential of both sides:

L = e^(-1) = 1/e

Great! We've found L. Now, we're given that -ln L = p. Substituting L = 1/e:

p = -ln(1/e) = -(-1) = 1

So, p = 1. Finally, we need to find the value of (1/(2p))^2. Plugging in p = 1:

(1/(2p))^2 = (1/(2*1))^2 = (1/2)^2 = 1/4

Conclusion: The Grand Finale

Guys, we did it! We successfully navigated the world of limits, factorials, Stirling's approximation, Riemann sums, and definite integrals to find the value of (1/(2p))^2. The answer is 1/4. This problem showcases the interconnectedness of different mathematical concepts and the power of using clever techniques to solve challenging problems.

I hope you enjoyed this mathematical journey as much as I did. Keep exploring, keep questioning, and keep learning! Math is an amazing adventure, and there's always something new to discover.

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Limit of nth Root of n Factorial Calculation and Solution

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If the limit as n approaches infinity of the nth root of (1 * 2 * 3 * ... * n) equals L, and -ln(L) = p, what is the value of (1/(2p))^2?