`T_(n) =sum _( r =2n )^(3n-1) (r)/(r ^(2) +n ^(2)), S_(n) = sum _(r =2n+1)^(3n) (r )/(r ^(2) + n ^(2)),` then ` AA n in {1,2,3...}:`

Options

A. `T_(n ) gt 1/2 ln 2`
B. ` S_(n) lt 1/2 ln 2`
C. ` T_(n) lt 1/2 ln 2`
D. ` S_(n)gt 1/2 ln 2`

Answer

To solve the problem, we need to analyze the sums \( T_n \) and \( S_n \) given by:

\[ T_n = \sum_{r=2n}^{3n-1} \frac{r}{r^2 + n^2} \]
\[ S_n = \sum_{r=2n+1}^{3n} \frac{r}{r^2 + n^2} \]

We will evaluate the limits of these sums as \( n \) approaches infinity.

### Step 1: Rewrite the sums in terms of \( n \)

We can express \( T_n \) and \( S_n \) in a more manageable form by factoring out \( n^2 \) from the denominator:

\[ T_n = \sum_{r=2n}^{3n-1} \frac{r}{n^2 \left( \frac{r^2}{n^2} + 1 \right)} = \frac{1}{n^2} \sum_{r=2n}^{3n-1} \frac{r}{\left( \frac{r}{n} \right)^2 + 1} \]

Similarly for \( S_n \):

\[ S_n = \sum_{r=2n+1}^{3n} \frac{r}{n^2 \left( \frac{r^2}{n^2} + 1 \right)} = \frac{1}{n^2} \sum_{r=2n+1}^{3n} \frac{r}{\left( \frac{r}{n} \right)^2 + 1} \]

### Step 2: Change of variables

Let \( x = \frac{r}{n} \). Then, as \( r \) goes from \( 2n \) to \( 3n-1 \), \( x \) goes from \( 2 \) to \( 3 - \frac{1}{n} \). The sum can be approximated by an integral as \( n \to \infty \):

\[ T_n \approx \frac{1}{n^2} \int_{2}^{3} \frac{nx}{x^2 + 1} \, n \, dx = \int_{2}^{3} \frac{x}{x^2 + 1} \, dx \]

### Step 3: Evaluate the integral

Now we compute the integral:

\[ \int \frac{x}{x^2 + 1} \, dx \]

Using substitution \( u = x^2 + 1 \), \( du = 2x \, dx \):

\[ \int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \ln |x^2 + 1| + C \]

Evaluating from \( 2 \) to \( 3 \):

\[ \left[ \frac{1}{2} \ln |x^2 + 1| \right]_{2}^{3} = \frac{1}{2} \left( \ln(10) - \ln(5) \right) = \frac{1}{2} \ln \left( \frac{10}{5} \right) = \frac{1}{2} \ln(2) \]

Thus, we find:

\[ T_n \to \frac{1}{2} \ln(2) \quad \text{as } n \to \infty \]

### Step 4: Evaluate \( S_n \)

Using similar steps for \( S_n \):

\[ S_n \approx \frac{1}{n^2} \int_{2}^{3} \frac{nx}{x^2 + 1} \, n \, dx = \int_{2}^{3} \frac{x}{x^2 + 1} \, dx \]

The evaluation is the same, leading to:

\[ S_n \to \frac{1}{2} \ln(2) \quad \text{as } n \to \infty \]

### Conclusion

Both \( T_n \) and \( S_n \) converge to \( \frac{1}{2} \ln(2) \) as \( n \) approaches infinity. Thus, we conclude:

\[ T_n > \frac{1}{2} \ln(2) \quad \text{and} \quad S_n > \frac{1}{2} \ln(2) \]