For all `n in N, 1 + 1/(sqrt(2))+1/(sqrt(3))+1/(sqrt(4))++1/(sqrt(n))`

Options

A. equal to `sqrt(n)`
B. less than or equal to `sqrt(n)`
C. greater tha or equal to `sqrt(n)`
D. none of these

Answer

To solve the problem, we need to analyze the series \( S(n) = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + \ldots + \frac{1}{\sqrt{n}} \) and determine its relationship with \( \sqrt{n} \).

### Step-by-step Solution:

1. **Understanding the Terms**:
We have the series \( S(n) = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} \). Each term in the series is of the form \( \frac{1}{\sqrt{k}} \) where \( k \) ranges from 1 to \( n \).

**Hint**: Identify the pattern in the terms of the series.

2. **Comparing Terms**:
Notice that \( \frac{1}{\sqrt{k}} \) is a decreasing function as \( k \) increases. This means that:
\[ \frac{1}{\sqrt{1}} > \frac{1}{\sqrt{2}} > \frac{1}{\sqrt{3}} > \ldots > \frac{1}{\sqrt{n}} \]
Therefore, each term \( \frac{1}{\sqrt{k}} \) is greater than or equal to \( \frac{1}{\sqrt{n}} \).

**Hint**: Use the property of decreasing functions to establish inequalities.

3. **Summing the Terms**:
Since there are \( n \) terms in the series, we can write:
\[ S(n) = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} \geq n \cdot \frac{1}{\sqrt{n}} = \sqrt{n} \]
This implies that the sum \( S(n) \) is greater than or equal to \( \sqrt{n} \).

**Hint**: Relate the number of terms to the minimum value of each term.

4. **Conclusion**:
We have established that:
\[ S(n) \geq \sqrt{n} \]
Therefore, the correct answer is that the sum \( S(n) \) is greater than or equal to \( \sqrt{n} \).

**Hint**: Verify your conclusion by checking the inequality derived.

### Final Answer:
The series \( S(n) = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} \) is greater than or equal to \( \sqrt{n} \).