Let `S_n=1/1^2 + 1/2^2 + 1/3^2` +….. + `1/n^2` and `T_n=2 -1/n` , then :

Options

A. `S_2 lt T_2`
B. If `S_k lt T_k` then `S_(k+1) lt T_(k+1)`
C. `S_n lt T_n AA n ge 2`
D. `S_n gt T_n AA n ge 2007`

Answer

To solve the problem, we need to analyze the sequences \( S_n \) and \( T_n \) defined as follows:

- \( S_n = \sum_{k=1}^{n} \frac{1}{k^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots + \frac{1}{n^2} \)
- \( T_n = 2 - \frac{1}{n} \)

We need to determine the relationship between \( S_n \) and \( T_n \) for various values of \( n \).

### Step 1: Calculate \( S_2 \) and \( T_2 \)

First, we calculate \( S_2 \):
\[ S_2 = \frac{1}{1^2} + \frac{1}{2^2} = 1 + \frac{1}{4} = \frac{5}{4} \]

Next, we calculate \( T_2 \):
\[ T_2 = 2 - \frac{1}{2} = 2 - 0.5 = \frac{3}{2} \]

### Step 2: Compare \( S_2 \) and \( T_2 \)

Now, we compare \( S_2 \) and \( T_2 \):
\[ S_2 = \frac{5}{4} = 1.25 \quad \text{and} \quad T_2 = \frac{3}{2} = 1.5 \]
Since \( S_2 < T_2 \), we conclude that the first option is correct.

### Step 3: Assume \( S_k < T_k \) and check for \( S_{k+1} < T_{k+1} \)

Assume \( S_k < T_k \), which means:
\[ S_k = \sum_{j=1}^{k} \frac{1}{j^2} < 2 - \frac{1}{k} \]

Now, we need to check if \( S_{k+1} < T_{k+1} \):
\[ S_{k+1} = S_k + \frac{1}{(k+1)^2} \]
\[ T_{k+1} = 2 - \frac{1}{k+1} \]

Substituting the assumption:
\[ S_{k+1} < 2 - \frac{1}{k} + \frac{1}{(k+1)^2} \]

### Step 4: Simplifying the inequality

We need to show:
\[ 2 - \frac{1}{k} + \frac{1}{(k+1)^2} < 2 - \frac{1}{k+1} \]

This simplifies to:
\[ -\frac{1}{k} + \frac{1}{(k+1)^2} < -\frac{1}{k+1} \]

Rearranging gives:
\[ \frac{1}{(k+1)^2} < \frac{1}{k} - \frac{1}{k+1} \]

### Step 5: Finding a common denominator

The right side can be simplified:
\[ \frac{1}{k} - \frac{1}{k+1} = \frac{(k+1) - k}{k(k+1)} = \frac{1}{k(k+1)} \]

Now we need to check:
\[ \frac{1}{(k+1)^2} < \frac{1}{k(k+1)} \]

Cross-multiplying gives:
\[ k < k + 1 \]
This is always true, so \( S_{k+1} < T_{k+1} \) holds.

### Step 6: Conclusion

By mathematical induction, we conclude that \( S_n < T_n \) for all \( n \geq 2 \).

### Final Answer

- Option A: Correct
- Option B: Correct
- Option C: Correct
- Option D: Incorrect