`lim_(n to oo) (1+(1+(1)/(2)+………+(1)/(n))/(n^(2)))^(n)` is equal to :

Options

A. `(1)/(2)`
B. 0
C. `(1)/(e )`
D. 1

Answer

To solve the limit

\[ \lim_{n \to \infty} \left( 1 + \frac{1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}}{n^2} \right)^n, \]

we can follow these steps:

### Step 1: Analyze the expression inside the limit
The expression inside the limit can be rewritten as:

\[ \lim_{n \to \infty} \left( 1 + \frac{H_n}{n^2} \right)^n, \]

where \( H_n = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} \) is the \( n \)-th harmonic number.

### Step 2: Approximate the harmonic number
The harmonic number \( H_n \) can be approximated as:

\[ H_n \sim \ln(n) + \gamma, \]

where \( \gamma \) is the Euler-Mascheroni constant. Thus, we can say:

\[ \frac{H_n}{n^2} \sim \frac{\ln(n)}{n^2}. \]

### Step 3: Substitute the approximation into the limit
Now, substituting this approximation into our limit gives:

\[ \lim_{n \to \infty} \left( 1 + \frac{\ln(n)}{n^2} \right)^n. \]

### Step 4: Identify the form of the limit
As \( n \to \infty \), \( \frac{\ln(n)}{n^2} \to 0 \). Therefore, we have a limit of the form \( (1 + 0)^n \), which is \( 1^\infty \).

### Step 5: Use the exponential limit form
To evaluate this limit, we can use the fact that:

\[ \lim_{n \to \infty} \left( 1 + \frac{a_n}{n} \right)^n = e^{\lim_{n \to \infty} a_n}, \]

where \( a_n = n \cdot \frac{\ln(n)}{n^2} = \frac{\ln(n)}{n} \).

### Step 6: Evaluate the limit of \( a_n \)
Now we need to evaluate:

\[ \lim_{n \to \infty} \frac{\ln(n)}{n}. \]

As \( n \to \infty \), \( \frac{\ln(n)}{n} \to 0 \).

### Step 7: Final evaluation
Thus, we have:

\[ \lim_{n \to \infty} \left( 1 + \frac{\ln(n)}{n^2} \right)^n = e^0 = 1. \]

### Conclusion
Therefore, the final result is:

\[ \lim_{n \to \infty} \left( 1 + \frac{H_n}{n^2} \right)^n = 1. \]