Step 1: Concept Overview:
The sequence \(b_n\) is defined with products and powers of \(n\), suggesting the use of a Riemann sum (definite integral) for limit evaluation. This often involves taking the logarithm. Note that the limit of \(b_n\) as given tends to 0. Since none of the answer choices are 0, we assume the problem intended to find \(\lim_{n \to \infty} (b_n)^{1/n}\), which we will now solve.
Step 2: Approach Summary:
1. Let \(L = \lim_{n \to \infty} (b_n)^{1/n}\).
2. Apply the natural logarithm: \(\ln(L) = \lim_{n \to \infty} \frac{1}{n}\ln(b_n)\).
3. Transform \(\ln(b_n)\) into a Riemann sum of the form \(\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})\).
4. Evaluate the definite integral: \(\int_{0}^{1}f(x)dx\).
Step 3: Detailed Solution:
Find \(L = \lim_{n \to \infty} (b_n)^{1/n}\), where
\[ b_n = \frac{n^n}{(n+1)(n+2)...(n+n)} = \frac{n^n}{\prod_{k=1}^{n}(n+k)} \].
Taking the n-th root:
\[ (b_n)^{1/n} = \left(\frac{n^n}{\prod_{k=1}^{n}(n+k)}\right)^{1/n} = \frac{n}{\left(\prod_{k=1}^{n}(n+k)\right)^{1/n}} \]
Taking the natural logarithm:
\[ \ln((b_n)^{1/n}) = \ln(n) - \frac{1}{n} \ln\left(\prod_{k=1}^{n}(n+k)\right) \]
\[ = \ln(n) - \frac{1}{n} \sum_{k=1}^{n} \ln(n+k) \]
Rewrite \(\ln(n)\) as \(\frac{1}{n} \sum_{k=1}^{n} \ln(n)\).
\[ \ln((b_n)^{1/n}) = \frac{1}{n} \sum_{k=1}^{n} \ln(n) - \frac{1}{n} \sum_{k=1}^{n} \ln(n+k) = \frac{1}{n} \sum_{k=1}^{n} (\ln(n) - \ln(n+k)) \]
\[ = \frac{1}{n} \sum_{k=1}^{n} \ln\left(\frac{n}{n+k}\right) = \frac{1}{n} \sum_{k=1}^{n} \ln\left(\frac{1}{1+k/n}\right) = -\frac{1}{n} \sum_{k=1}^{n} \ln\left(1+\frac{k}{n}\right) \]
As \(n \to \infty\), this becomes a Riemann sum for \(f(x) = \ln(1+x)\) on \([0, 1]\).
\[ \lim_{n \to \infty} \ln((b_n)^{1/n}) = -\int_{0}^{1} \ln(1+x) dx \]
Using integration by parts, \(\int u dv = uv - \int v du\), with \(u = \ln(1+x)\) and \(dv = dx\), so \(du = \frac{1}{1+x}dx\) and \(v = x\).
\[ \int \ln(1+x) dx = x\ln(1+x) - \int \frac{x}{1+x} dx = x\ln(1+x) - \int \frac{1+x-1}{1+x} dx \]
\[ = x\ln(1+x) - \int \left(1 - \frac{1}{1+x}\right) dx = x\ln(1+x) - (x - \ln(1+x)) = (x+1)\ln(1+x) - x \]
Evaluating the definite integral:
\[ \int_{0}^{1} \ln(1+x) dx = [(x+1)\ln(1+x) - x]_{0}^{1} = ((2)\ln(2) - 1) - ((1)\ln(1) - 0) = 2\ln(2) - 1 = \ln(4) - \ln(e) = \ln(4/e) \]
Thus, \(\lim_{n \to \infty} \ln((b_n)^{1/n}) = -\ln(4/e) = \ln((4/e)^{-1}) = \ln(e/4)\).
Therefore, \(L = \lim_{n \to \infty} (b_n)^{1/n} = e/4\).
Step 4: Solution:
If the problem intended to find the limit of \((b_n)^{1/n}\), the answer is \(e/4\).