What is the limiting value of the expression
`({(n +1) (n+2).....(2n)} ^(1/n))/(n) ` when n tends to infinity ?

Options

A. `1/2`
B. e
C. `2/e`
D. `4/e`

Answer

To find the limiting value of the expression

\[ \frac{(n + 1)(n + 2) \cdots (2n)^{1/n}}{n} \]

as \( n \) tends to infinity, we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression in a more manageable form. The product \( (n + 1)(n + 2) \cdots (2n) \) can be expressed as:

\[ \frac{(2n)!}{n!} \]

Thus, our expression becomes:

\[ \frac{((2n)! / n!)^{1/n}}{n} \]

### Step 2: Simplify the expression
Now we can rewrite our expression as:

\[ \frac{(2n)!^{1/n}}{(n!)^{1/n} \cdot n} \]

### Step 3: Use Stirling's approximation
For large \( n \), we can use Stirling's approximation, which states:

\[ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \]

Applying this to both \( (2n)! \) and \( n! \):

\[ (2n)! \sim \sqrt{4 \pi n} \left(\frac{2n}{e}\right)^{2n} \]
\[ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \]

### Step 4: Substitute Stirling's approximation
Substituting these approximations into our expression gives:

\[ \frac{\left(\sqrt{4 \pi n} \left(\frac{2n}{e}\right)^{2n}\right)^{1/n}}{\left(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^{n}\right)^{1/n} \cdot n} \]

### Step 5: Simplify further
This simplifies to:

\[ \frac{\sqrt{4 \pi n}^{1/n} \cdot \left(\frac{2n}{e}\right)^{2}}{\sqrt{2 \pi n}^{1/n} \cdot \left(\frac{n}{e}\right) \cdot n} \]

As \( n \to \infty \), \( \sqrt{4 \pi n}^{1/n} \) and \( \sqrt{2 \pi n}^{1/n} \) both approach 1. Thus, we can focus on the main terms:

\[ \frac{4}{1} = 4 \]

### Step 6: Final limit
Thus, we have:

\[ \lim_{n \to \infty} \frac{(n + 1)(n + 2) \cdots (2n)^{1/n}}{n} = \frac{4}{1} = 4 \]

### Conclusion
The limiting value of the expression is:

\[ \frac{4}{e} \]