The vlaue of `lim_(n rarroo) sum_(k=1)^n frac{n^3}{(n^2+k^2)(n^2+3k^2)` is

Options

A. `frac{pi}{2sqrt2} - frac{pi}{4}`
B. `frac{pi}{2sqrt3} - frac{pi}{8}`
C. `frac{pi}{2sqrt3} + frac{pi}{4}`
D. `frac{pi}{sqrt3} - frac{pi}{8}`

Answer

To solve the limit

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

we will follow these steps:

### Step 1: Rewrite the expression

We start by rewriting the summation:

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

### Step 2: Factor out \(n^4\)

Factoring out \(n^4\) from the denominator gives us:

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

### Step 3: Convert the sum to an integral

As \(n \to \infty\), we can replace the sum with an integral. The term \(\frac{1}{n}\) can be interpreted as \(dx\) where \(x = \frac{k}{n}\). The limits of integration will be from \(0\) to \(1\) as \(k\) goes from \(1\) to \(n\).

Thus, we have:

\[ \sum_{k=1}^{n} \frac{1}{n \left(1 + \frac{k^2}{n^2}\right)\left(1 + 3\frac{k^2}{n^2}\right)} \approx \int_{0}^{1} \frac{1}{(1 + x^2)(1 + 3x^2)} \, dx. \]

### Step 4: Solve the integral using partial fractions

To evaluate the integral

\[ \int_{0}^{1} \frac{1}{(1 + x^2)(1 + 3x^2)} \, dx, \]

we will 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 gives:

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

### Step 5: Solve for coefficients A and B

Setting \(x^2 = 0\):

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

Setting \(x^2 = -\frac{1}{3}\):

\[ 1 = A(1) + B(0) \implies A = 1. \]

Substituting \(A = 1\) into \(A + B = 1\):

\[ 1 + B = 1 \implies B = 0. \]

Thus, we have:

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

### Step 6: Integrate the partial fractions

Now we can integrate:

\[ \int_{0}^{1} \left( \frac{1}{1 + x^2} - \frac{1}{3(1 + 3x^2)} \right) dx. \]

The first integral is:

\[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x). \]

The second integral is:

\[ \int \frac{1}{3(1 + 3x^2)} \, dx = \frac{1}{3} \cdot \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}x). \]

### Step 7: Evaluate the definite integral

Evaluating from \(0\) to \(1\):

\[ \left[ \tan^{-1}(x) - \frac{1}{3\sqrt{3}} \tan^{-1}(\sqrt{3}x) \right]_{0}^{1}. \]

At \(x = 1\):

\[ \tan^{-1}(1) - \frac{1}{3\sqrt{3}} \tan^{-1}(\sqrt{3}) = \frac{\pi}{4} - \frac{1}{3\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi}{4} - \frac{\pi}{9\sqrt{3}}. \]

At \(x = 0\):

\[ \tan^{-1}(0) - 0 = 0. \]

### Final Result

Thus, the limit evaluates to:

\[ \frac{\pi}{4} - \frac{\pi}{9\sqrt{3}}. \]