Evaluate Limit: N(nth Root((n+1)...(2n)) - N)

Hey guys! Today, we're diving deep into a fascinating calculus problem: evaluating the limit of a somewhat intimidating expression. Specifically, we're tackling this bad boy:

$$\lim_{n\to\infty}n(\sqrt[n]{(n+1)(n+2)(n+3)...(2n)}-n)$$

This isn't your run-of-the-mill limit, so buckle up! We'll break it down step by step, ensuring you grasp every nuance along the way. Let's make calculus fun, shall we?

Initial Thoughts and Challenges

At first glance, this limit looks daunting. The nth root of a product of n terms, all multiplied by n, and then we subtract n? Yikes! A direct substitution of infinity will lead to an indeterminate form, so we need a clever approach. It's essential to recognize that standard limit techniques might fall short here, urging us to explore more sophisticated methods. The initial instinct might be to try L'Hôpital's Rule, but that could get messy real quick. Instead, we'll aim for something more elegant and insightful.

Many might initially think about using the AM-GM inequality, which you know, states that the Arithmetic Mean is always greater than or equal to the Geometric Mean. While a valid idea, directly applying AM-GM might not lead us to the tightest bounds needed to evaluate this limit effectively. We need to manipulate the expression to reveal its true behavior as n approaches infinity.

The Logarithmic Transformation

A brilliant strategy to simplify expressions involving products and roots is to employ logarithms. Let's define:

$$L = \lim_{n\to\infty}n(\sqrt[n]{(n+1)(n+2)(n+3)...(2n)}-n)$$

We'll focus on the term inside the limit and introduce a logarithmic transformation. Let

$$a_n = \sqrt[n]{(n+1)(n+2)...(2n)}$$

Then

$$\ln(a_n) = \frac{1}{n} \ln[(n+1)(n+2)...(2n)] = \frac{1}{n} \sum_{k=1}^{n} \ln(n+k)$$

Now, we can rewrite this as:

$$\ln(a_n) = \frac{1}{n} \sum_{k=1}^{n} \ln[n(1+\frac{k}{n})] = \frac{1}{n} \sum_{k=1}^{n} [\ln(n) + \ln(1+\frac{k}{n})] = \ln(n) + \frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n})$$

This is where things start to get interesting. Notice that the sum looks like a Riemann sum!

Riemann Sums to the Rescue

The expression $\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n})$ is indeed a Riemann sum that converges to a definite integral. As n approaches infinity, this sum approaches:

$$\int_{0}^{1} \ln(1+x) dx$$

We can evaluate this integral using integration by parts. Let $u = \ln(1+x)$ and $dv = dx$. Then $du = \frac{1}{1+x} dx$ and $v = x$. Applying integration by parts:

$$\int_{0}^{1} \ln(1+x) dx = [x\ln(1+x)]_{0}^{1} - \int_{0}^{1} \frac{x}{1+x} dx = \ln(2) - \int_{0}^{1} (1 - \frac{1}{1+x}) dx$$

$$= \ln(2) - [x - \ln(1+x)]_{0}^{1} = \ln(2) - (1 - \ln(2)) = 2\ln(2) - 1$$

Thus,

$$\lim_{n\to\infty} \ln(a_n) = \ln(n) + 2\ln(2) - 1$$

Now we exponentiate to find the limit of $a_n$:

$$\lim_{n\to\infty} a_n = \lim_{n\to\infty} e^{\ln(n) + 2\ln(2) - 1} = \lim_{n\to\infty} n e^{2\ln(2) - 1} = \lim_{n\to\infty} n e^{\ln(4) - 1} = \frac{4}{e}n$$

Back to the Original Limit

Now we return to our original limit:

$$L = \lim_{n\to\infty} n(a_n - n) = \lim_{n\to\infty} n(\frac{4}{e}n - n) = \lim_{n\to\infty} n^2(\frac{4}{e} - 1)$$

Oh no! This goes to infinity. We made a mistake somewhere. Let's re-examine our steps. Specifically, the approximation of the sum with an integral might not be accurate enough.

A More Refined Approach: Stirling's Approximation

Since the Riemann sum approach didn't give us the precision we need, let's try something more powerful: Stirling's approximation. Stirling's formula provides a very accurate approximation for the factorial function, and it's particularly useful when dealing with large n. Recall Stirling's approximation:

$$n! \approx \sqrt{2\pi n} (\frac{n}{e})^n$$

We can rewrite the expression inside the nth root as:

$$\sqrt[n]{(n+1)(n+2)...(2n)} = \sqrt[n]{\frac{(2n)!}{n!}}$$

Now, applying Stirling's approximation:

$$\frac{(2n)!}{n!} \approx \frac{\sqrt{4\pi n} (\frac{2n}{e})^{2n}}{\sqrt{2\pi n} (\frac{n}{e})^n} = \sqrt{2} \frac{(2n)^{2n}}{n^n e^n} = \sqrt{2} \frac{4^n n^{2n}}{n^n e^n} = \sqrt{2} (\frac{4n}{e})^n$$

Taking the nth root:

$$\sqrt[n]{\frac{(2n)!}{n!}} \approx (\sqrt{2})^{\frac{1}{n}} \frac{4n}{e}$$

As $n \to \infty$, $(\sqrt{2})^{\frac{1}{n}} \to 1$. Thus,

$$\lim_{n\to\infty} \sqrt[n]{\frac{(2n)!}{n!}} = \frac{4n}{e}$$

Now we revisit the original limit:

$$L = \lim_{n\to\infty} n(\sqrt[n]{(n+1)(n+2)...(2n)}-n) = \lim_{n\to\infty} n(\frac{4n}{e} - n) = \lim_{n\to\infty} n^2(\frac{4}{e} - 1)$$

Again, we face the same issue – the limit goes to infinity. This indicates that our approximation, while powerful, might still not be precise enough to capture the subtle difference between the nth root and n. We need an even more refined approach.

Time for Taylor Series Expansion

Since we are dealing with limits as n approaches infinity, and we've already tried Riemann sums and Stirling's approximation, let's try to express our functions using Taylor series. We have already established that:

$$\ln(a_n) = \ln(n) + \frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n})$$

Let's consider the Taylor expansion of $\ln(1+x)$ around $x=0$:

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$$

So,

$$\ln(1+\frac{k}{n}) = \frac{k}{n} - \frac{k^2}{2n^2} + \frac{k^3}{3n^3} - ...$$

$$\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{1}{n} \sum_{k=1}^{n} (\frac{k}{n} - \frac{k^2}{2n^2} + \frac{k^3}{3n^3} - ...) = \frac{1}{n^2} \sum_{k=1}^{n} k - \frac{1}{2n^3} \sum_{k=1}^{n} k^2 + \frac{1}{3n^4} \sum_{k=1}^{n} k^3 - ...$$

Using the formulas for the sum of powers of integers:

\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$, $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum_{k=1}^{n} k^3 = (\frac{n(n+1)}{2})^2

$$\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{1}{n^2} \frac{n(n+1)}{2} - \frac{1}{2n^3} \frac{n(n+1)(2n+1)}{6} + \frac{1}{3n^4} (\frac{n(n+1)}{2})^2 - ...$$

$$\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{n^2 + n}{2n^2} - \frac{2n^3 + 3n^2 + n}{12n^3} + \frac{n^4 + 2n^3 + n^2}{12n^4} - ...$$

$$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{1}{2} - \frac{2}{12} + \frac{1}{12} = \frac{1}{2} - \frac{1}{6} + \frac{1}{12} = \frac{6-2+1}{12} = \frac{5}{12}$$

Therefore,

$$\lim_{n\to\infty} \ln(a_n) = \ln(n) + \frac{5}{12}$$

$$a_n = e^{\ln(n) + \frac{5}{12}} = n e^{\frac{5}{12}}$$

Then

$$\lim_{n\to\infty} n(a_n - n) = \lim_{n\to\infty} n(ne^{\frac{5}{12}}-n) = \lim_{n\to\infty} n^2(e^{\frac{5}{12}}-1) = \infty$$

Still infinity! Where did we go wrong AGAIN?! Let's slow down and reassess. The Taylor series expansion seems correct, but the problem might be that the higher order terms, though smaller, still contribute significantly when multiplied by $n^2$.

The Final Piece: A More Precise Taylor Expansion

Okay, guys, let's get serious. The trick here is to use a more precise Taylor expansion and keep track of the error terms. We have:

$$\ln(a_n) = \ln(n) + \frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n})$$

Using the Taylor expansion $\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$:

$$\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{1}{n} \sum_{k=1}^{n} (\frac{k}{n} - \frac{k^2}{2n^2} + O(\frac{k^3}{n^3}))$$

$$\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{1}{n^2} \sum_{k=1}^{n} k - \frac{1}{2n^3} \sum_{k=1}^{n} k^2 + O(\frac{1}{n^4} \sum_{k=1}^{n} k^3)$$

$$\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{n(n+1)}{2n^2} - \frac{n(n+1)(2n+1)}{6(2n^3)} + O(\frac{n^2(n+1)^2}{4n^4})$$

$$\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{1}{2} + \frac{1}{2n} - \frac{1}{6}(2 + \frac{3}{n} + \frac{1}{n^2}) + O(\frac{1}{n^2})$$

$$\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n}) = \frac{1}{2} - \frac{1}{3} + \frac{1}{2n} - \frac{1}{2n} + O(\frac{1}{n^2}) = \frac{1}{6} + O(\frac{1}{n^2})$$

Therefore,

$$\ln(a_n) = \ln(n) + \frac{1}{6} + O(\frac{1}{n^2})$$

$$a_n = n e^{\frac{1}{6} + O(\frac{1}{n^2})} = n e^{\frac{1}{6}} (1 + O(\frac{1}{n^2}))$$

$$a_n = n e^{\frac{1}{6}} + O(\frac{1}{n})$$

Now, let's plug this back into our limit:

$$L = \lim_{n\to\infty} n(a_n - n) = \lim_{n\to\infty} n(n e^{\frac{1}{6}} + O(\frac{1}{n}) - n) = \lim_{n\to\infty} n^2(e^{\frac{1}{6}} - 1) + O(1)$$

We made another mistake! Let's go back to the formula:

$$\ln(a_n) = \ln(n) + \frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n})$$

Then:

$$a_n = n e^{\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n})}$$

$$a_n - n = n(e^{\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n})} - 1)$$

Using $e^x - 1 \approx x$ for small $x$:

$$a_n - n \approx n(\frac{1}{n} \sum_{k=1}^{n} \ln(1+\frac{k}{n})) = \sum_{k=1}^{n} \ln(1+\frac{k}{n})$$

Then:

$$\lim_{n\to\infty} n(a_n - n) = \lim_{n\to\infty} n(\frac{4n}{e} - n) = \lim_{n\to\infty} \frac{4}{e}n^2 - n^2$$

Final Solution

Let $a_n = \sqrt[n]{(n+1)(n+2)...(2n)}$. Then $\ln(a_n) = \frac{1}{n} \sum_{k=1}^n \ln(n+k) = \frac{1}{n} \sum_{k=1}^n \ln(n(1+\frac{k}{n})) = \frac{1}{n} \sum_{k=1}^n (\ln(n) + \ln(1+\frac{k}{n})) = \ln(n) + \frac{1}{n} \sum_{k=1}^n \ln(1+\frac{k}{n})$. As $n \to \infty, \frac{1}{n} \sum_{k=1}^n \ln(1+\frac{k}{n}) \to \int_0^1 \ln(1+x)dx = (1+x)\ln(1+x) - x |_0^1 = 2\ln(2) - 1 = \ln(4) - 1$. So, $a_n \approx e^{\ln(n) + \ln(4) - 1} = \frac{4n}{e}$. Then, $n(a_n - n) = n(\frac{4n}{e} - n) = n^2(\frac{4}{e} - 1)$. Now we need a better approximation: $a_n = \frac{4n}{e} + c + O(\frac{1}{n})$. Let $L = \lim_{n\to\infty} n(a_n - n)$. Then $a_n = \frac{4n}{e} + \frac{c}{n} + ...$. We look for the limit of $\frac{1}{n} \sum_{k=1}^n \ln(1+\frac{k}{n})$. $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - ...$. So, $\frac{1}{n} \sum_{k=1}^n (\frac{k}{n} - \frac{k^2}{2n^2} + \frac{k^3}{3n^3} - ...) = \frac{1}{n^2} \frac{n(n+1)}{2} - \frac{1}{2n^3} \frac{n(n+1)(2n+1)}{6} + \frac{1}{3n^4} (\frac{n(n+1)}{2})^2 - ... = (\frac{1}{2} + \frac{1}{2n}) - \frac{1}{12} (2 + \frac{3}{n} + \frac{1}{n^2}) + \frac{1}{12} (1 + \frac{2}{n} + \frac{1}{n^2}) - ... \approx \frac{1}{2} - \frac{1}{6} + \frac{1}{12} = \frac{6-2+1}{12} = \frac{5}{12}$. So, $a_n = n e^{\frac{5}{12}}$. Then $L = \lim_{n\to\infty} n^2(e^{\frac{5}{12}} - 1) = \infty$. However, we need to find the next term in the expansion. We've found that $\ln(a_n) = \ln(n) + \frac{5}{12} + O(\frac{1}{n})$. This implies $a_n = n \exp(\frac{5}{12} + O(\frac{1}{n})) \approx n e^{\frac{5}{12}}(1 + O(\frac{1}{n}))$. Then $L = \lim_{n\to\infty} n(n e^{\frac{5}{12}} - n + O(1)) = \lim_{n\to\infty} n^2(e^{\frac{5}{12}} - 1) + ...$. Clearly we are missing a term, since $n(\sqrt[n]{(n+1)(n+2)(n+3)...(2n)}-n)$ approaches infinity as $n$ goes to infinity.

Conclusion

Wow, that was a journey! We explored various techniques, from logarithmic transformations and Riemann sums to Stirling's approximation and Taylor series expansions. While we didn't arrive at a finite limit using these methods, we gained valuable insights into the behavior of the expression as n approaches infinity. The limit seems to diverge, but a more precise approach might be needed for a definitive answer. This problem highlights the importance of choosing the right tool for the job and the power of combining different mathematical techniques to tackle complex problems. Keep exploring, guys, and never stop questioning!