The value of `lim_(n rarr oo)sum_(k=1)^(n)(n^(3))/((n^(2)+k^(2))(n^(2)+3k^(2)))` is:

Options

A. `(13(2sqrt(3)-3)pi)/(8)`
B. `((2sqrt(3)+3)pi)/(24)`
C. `(pi)/(8(2sqrt(3)+3))`
D. `(13 pi)/(8(4sqrt(3)+3))`

Answer

To solve the limit problem

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^3}{(n^2 + k^2)(n^2 + 3k^2)}, \]

we will follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the limit expression. Notice that both terms in the denominator contain \(n^2\). We can factor \(n^2\) out of each term in the denominator:

\[ = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^3}{n^2 \left(1 + \frac{k^2}{n^2}\right) \cdot n^2 \left(1 + \frac{3k^2}{n^2}\right)}. \]

This simplifies to:

\[ = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^3}{n^4 \left(1 + \frac{k^2}{n^2}\right) \left(1 + \frac{3k^2}{n^2}\right)} = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n \left(1 + \frac{k^2}{n^2}\right) \left(1 + \frac{3k^2}{n^2}\right)}. \]

### Step 2: Change of variables
Next, we can make a substitution to convert the sum into a Riemann integral. Let \(x = \frac{k}{n}\). Then, \(k = nx\) and \(dk = n \, dx\). The limits of \(k\) from \(1\) to \(n\) correspond to \(x\) from \(\frac{1}{n}\) to \(1\). Thus, we can rewrite the sum as:

\[ \sum_{k=1}^{n} \frac{1}{n \left(1 + \frac{k^2}{n^2}\right) \left(1 + \frac{3k^2}{n^2}\right)} \approx \int_0^1 \frac{1}{\left(1 + x^2\right) \left(1 + 3x^2\right)} \, dx. \]

### Step 3: Evaluate the integral
Now we need to evaluate the integral:

\[ \int_0^1 \frac{1}{(1 + x^2)(1 + 3x^2)} \, dx. \]

We can use partial fraction decomposition:

\[ \frac{1}{(1 + x^2)(1 + 3x^2)} = \frac{A}{1 + x^2} + \frac{B}{1 + 3x^2}. \]

Multiplying through by the denominator \((1 + x^2)(1 + 3x^2)\) gives:

\[ 1 = A(1 + 3x^2) + B(1 + x^2). \]

Expanding and equating coefficients, we can solve for \(A\) and \(B\).

### Step 4: Solve for coefficients
Setting \(x = 0\):

\[ 1 = A + B \implies A + B = 1. \]

Setting \(x^2 = 1\):

\[ 1 = A(1 + 3) + B(1 + 1) \implies 4A + 2B = 1. \]

Now we have the system of equations:

1. \(A + B = 1\)
2. \(4A + 2B = 1\)

From the first equation, \(B = 1 - A\). Substituting into the second equation:

\[ 4A + 2(1 - A) = 1 \implies 4A + 2 - 2A = 1 \implies 2A = -1 \implies A = \frac{1}{2}, B = \frac{1}{2}. \]

### Step 5: Integrate
Now we can rewrite the integral:

\[ \int_0^1 \left(\frac{1/2}{1 + x^2} + \frac{1/2}{1 + 3x^2}\right) \, dx = \frac{1}{2} \int_0^1 \frac{1}{1 + x^2} \, dx + \frac{1}{2} \int_0^1 \frac{1}{1 + 3x^2} \, dx. \]

The first integral evaluates to:

\[ \frac{1}{2} \left[\tan^{-1}(x)\right]_0^1 = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}. \]

The second integral evaluates to:

\[ \frac{1}{2} \left[\frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}x)\right]_0^1 = \frac{1}{2\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi}{6\sqrt{3}}. \]

### Final Step: Combine results
Combining both results gives:

\[ \frac{\pi}{8} + \frac{\pi}{6\sqrt{3}}. \]

### Conclusion
Thus, the final value of the limit is:

\[ \frac{\pi}{8} + \frac{\pi}{6\sqrt{3}}. \]