Question

The limit of the sequence,
\(\{b_n; b_n = \frac{n^n}{(n+1)(n+2)...(n+n)}; n>0\}\), is

• A. \(\frac{e}{2}\)
• B. \(\frac{e}{4}\)
• C. \(e\)
• D. \(\frac{1}{e}\)

Hint:

When a sequence's general term involves products or factorials, taking the logarithm and forming a Riemann sum is a standard technique. The key is to transform the expression into the form \(\lim_{n \to \infty} \frac{1}{n} \sum f(\frac{k}{n})\), which equals \(\int_0^1 f(x) dx\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The expression for the sequence \(b_n\) involves products and powers of \(n\), which is a common pattern for limits that can be evaluated by converting them into a definite integral (a Riemann sum). This is typically done by taking the logarithm of the expression. Note: The limit of \(b_n\) as written approaches 0. However, none of the options is 0, suggesting a common exam question typo where the limit of the n-th root, \(\lim_{n \to \infty} (b_n)^{1/n}\), is intended. We will solve this intended problem.

Step 2: Key Formula or Approach:
1. Let \(L = \lim_{n \to \infty} (b_n)^{1/n}\).
2. Take the natural logarithm: \(\ln(L) = \lim_{n \to \infty} \frac{1}{n}\ln(b_n)\).
3. Manipulate the expression for \(\ln(b_n)\) into the form of a Riemann sum: \(\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})\).
4. The limit of the Riemann sum is the definite integral: \(\int_{0}^{1}f(x)dx\).

Step 3: Detailed Explanation:
Let's find \(L = \lim_{n \to \infty} (b_n)^{1/n}\). \[ b_n = \frac{n^n}{(n+1)(n+2)...(n+n)} = \frac{n^n}{\prod_{k=1}^{n}(n+k)} \] Taking the n-th root: \[ (b_n)^{1/n} = \left(\frac{n^n}{\prod_{k=1}^{n}(n+k)}\right)^{1/n} = \frac{n}{\left(\prod_{k=1}^{n}(n+k)\right)^{1/n}} \] Let's take the natural logarithm of this expression: \[ \ln((b_n)^{1/n}) = \ln(n) - \frac{1}{n} \ln\left(\prod_{k=1}^{n}(n+k)\right) \] \[ = \ln(n) - \frac{1}{n} \sum_{k=1}^{n} \ln(n+k) \] We can write \(\ln(n)\) as \(\frac{1}{n} \sum_{k=1}^{n} \ln(n)\). \[ \ln((b_n)^{1/n}) = \frac{1}{n} \sum_{k=1}^{n} \ln(n) - \frac{1}{n} \sum_{k=1}^{n} \ln(n+k) = \frac{1}{n} \sum_{k=1}^{n} (\ln(n) - \ln(n+k)) \] \[ = \frac{1}{n} \sum_{k=1}^{n} \ln\left(\frac{n}{n+k}\right) = \frac{1}{n} \sum_{k=1}^{n} \ln\left(\frac{1}{1+k/n}\right) = -\frac{1}{n} \sum_{k=1}^{n} \ln\left(1+\frac{k}{n}\right) \] Now we take the limit as \(n \to \infty\). This expression is a Riemann sum for the function \(f(x) = \ln(1+x)\) over the interval \([0, 1]\). \[ \lim_{n \to \infty} \ln((b_n)^{1/n}) = -\int_{0}^{1} \ln(1+x) dx \] We evaluate the integral using integration by parts, \(\int u dv = uv - \int v du\). Let \(u = \ln(1+x)\) and \(dv = dx\). Then \(du = \frac{1}{1+x}dx\) 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: \[ \int_{0}^{1} \ln(1+x) dx = [(x+1)\ln(1+x) - x]_{0}^{1} = ((2)\ln(2) - 1) - ((1)\ln(1) - 0) = 2\ln(2) - 1 = \ln(4) - \ln(e) = \ln(4/e) \] So, \(\lim_{n \to \infty} \ln((b_n)^{1/n}) = -\ln(4/e) = \ln((4/e)^{-1}) = \ln(e/4)\). Therefore, \(L = \lim_{n \to \infty} (b_n)^{1/n} = e/4\).
Step 4: Final Answer:
Assuming the question intended to ask for the limit of \((b_n)^{1/n}\), the value is \(e/4\).