Given that
`lim_(nto oo) sum_(r=1)^(n) (log (r+n)-log n)/(n)=2(log 2-(1)/(2))`,
` lim_(n to oo) [(1)/(n^k)[(n+1)^k(n+2)^k.....(n+n)^k]]^(1//n)`, is

Options

A. `(4k)/(e)`
B. `ksqrt((4)/(e)`
C. `((4)/(e))^k`
D. `((e)/(4))^k`.

Answer

To solve the given limit problem step by step, we will analyze the expression and apply relevant mathematical principles.

### Step 1: Understanding the Limit Expression
We need to evaluate the limit:
\[ L = \lim_{n \to \infty} \left[ \frac{1}{n^k} \prod_{r=1}^{n} (n+r)^k \right]^{\frac{1}{n}} \]
This can be rewritten as:
\[ L = \lim_{n \to \infty} \left[ \frac{(n+1)^k (n+2)^k \cdots (n+n)^k}{n^k} \right]^{\frac{1}{n}} \]

### Step 2: Simplifying the Product
We can factor out \(n^k\) from each term in the product:
\[ = \lim_{n \to \infty} \left[ \frac{(n(1+\frac{1}{n}))^k (n(1+\frac{2}{n}))^k \cdots (n(1+\frac{n}{n}))^k}{n^k} \right]^{\frac{1}{n}} \]
This simplifies to:
\[ = \lim_{n \to \infty} \left[ n^{nk} \cdot (1+\frac{1}{n})^k (1+\frac{2}{n})^k \cdots (1+\frac{n}{n})^k \cdot \frac{1}{n^k} \right]^{\frac{1}{n}} \]
\[ = \lim_{n \to \infty} \left[ n^{nk - k} \cdot (1+\frac{1}{n})^k (1+\frac{2}{n})^k \cdots (1+\frac{n}{n})^k \right]^{\frac{1}{n}} \]
\[ = \lim_{n \to \infty} n^{k(n-1)/n} \cdot \left[(1+\frac{1}{n})^k (1+\frac{2}{n})^k \cdots (1+\frac{n}{n})^k \right]^{\frac{1}{n}} \]

### Step 3: Evaluating the Limit of the Product
Now we need to evaluate:
\[ \lim_{n \to \infty} \left[(1+\frac{1}{n})^k (1+\frac{2}{n})^k \cdots (1+\frac{n}{n})^k \right]^{\frac{1}{n}} \]
This can be expressed as:
\[ = \lim_{n \to \infty} \exp\left( \frac{1}{n} \sum_{r=1}^{n} k \log(1+\frac{r}{n}) \right) \]
Using the property of logarithms, we can rewrite this as:
\[ = \lim_{n \to \infty} \exp\left( k \cdot \frac{1}{n} \sum_{r=1}^{n} \log(1+\frac{r}{n}) \right) \]

### Step 4: Recognizing the Riemann Sum
As \(n \to \infty\), the sum can be recognized as a Riemann sum for the integral:
\[ \int_0^1 \log(1+x) \, dx \]
Calculating this integral:
\[ \int_0^1 \log(1+x) \, dx = [ (1+x) \log(1+x) - (1+x) ]_0^1 = [2 \log 2 - 2] - [0] = 2 \log 2 - 2 \]

### Step 5: Putting it All Together
Thus, we have:
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \log(1+\frac{r}{n}) = 2 \log 2 - 1 \]
Therefore, we can conclude:
\[ L = \exp\left(k(2 \log 2 - 1)\right) = \exp\left(2k \log 2 - k\right) = \frac{4^k}{e^k} \]
Thus, the final answer is:
\[ L = \left(\frac{4}{e}\right)^k \]