Question

The value of \(\lim_{{n \to \infty}} \sum_{{k=1}}^{n} \frac{n^3}{{(n^2 + k^2)(n^2 + 3k^2)}}\) is 

• A. \( \frac{13\pi}{8(4\sqrt{3} + 3)} \)
• B. \( \frac{(2\sqrt{3} + 3) \pi}{24} \)
• C. \( \frac{13(2\sqrt{3} - 3) \pi}{8} \)
• D. \( \frac{\pi}{8(2\sqrt{3} + 3)} \)

Answer 1

The Correct Option is A

The problem involves evaluating the limit of a summation, which can be transformed into a definite integral.

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

Concept Used:

The limit of a sum is equivalent to a definite integral, based on the Riemann sum definition:

\[\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(\frac{k}{n}\right) = \int_{0}^{1} f(x) \, dx\]

The resultant integral will be computed using partial fraction decomposition and the standard integral formula:

\[\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C\]

Step-by-Step Solution:

Step 1: Transform the limit of the sum into a definite integral.

The general term of the sum is manipulated to match the \( \frac{1}{n} f\left(\frac{k}{n}\right) \) format.

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

By comparison with the Riemann sum formula, we identify \( x = \frac{k}{n} \) and \( f(x) = \frac{1}{(1+x^2)(1+3x^2)} \). The limit is thus represented as:

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

Step 2: Apply partial fraction decomposition to the integrand.

Let \( y = x^2 \). The expression \( \frac{1}{(1+y)(1+3y)} \) is decomposed as:

\[\frac{1}{(1+y)(1+3y)} = \frac{A}{1+y} + \frac{B}{1+3y}\]

Multiplying by the denominator yields \( 1 = A(1+3y) + B(1+y) \).

Setting \( y = -1 \) gives \( 1 = A(1-3) \), so \( A = -\frac{1}{2} \).

Setting \( y = -1/3 \) gives \( 1 = B(1 - 1/3) \), so \( B = \frac{3}{2} \).

Substituting back \( y = x^2 \), the integrand is rewritten as:

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

Step 3: Compute the definite integral.

The integral is now:

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

Each integral is evaluated individually:

\[\int_{0}^{1} \frac{3}{1+(\sqrt{3}x)^2} \, dx = 3 \left[ \frac{1}{\sqrt{3}} \arctan(\sqrt{3}x) \right]_{0}^{1} = \sqrt{3} (\arctan(\sqrt{3}) - \arctan(0)) = \sqrt{3} \left( \frac{\pi}{3} \right) = \frac{\pi}{\sqrt{3}}\]\[\int_{0}^{1} \frac{1}{1+x^2} \, dx = [\arctan(x)]_{0}^{1} = \arctan(1) - \arctan(0) = \frac{\pi}{4}\]

Step 4: Combine the results to determine the value of L.

\[L = \frac{1}{2} \left[ \frac{\pi}{\sqrt{3}} - \frac{\pi}{4} \right] = \frac{\pi}{2} \left( \frac{4 - \sqrt{3}}{4\sqrt{3}} \right) = \frac{\pi (4 - \sqrt{3})}{8\sqrt{3}}\]

Rationalizing the denominator yields:

\[L = \frac{\pi (4 - \sqrt{3})}{8\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\pi (4\sqrt{3} - 3)}{24}\]

Final Computation & Result:

The limit's value is \( \frac{\pi (4\sqrt{3} - 3)}{24} \). We verify which given option matches this result. Consider option (2):

\[\frac{13\pi}{8(4\sqrt{3} + 3)}\]

Rationalizing its denominator:

\[\frac{13\pi}{8(4\sqrt{3} + 3)} \times \frac{4\sqrt{3} - 3}{4\sqrt{3} - 3} = \frac{13\pi (4\sqrt{3} - 3)}{8 \left( (4\sqrt{3})^2 - 3^2 \right)}\]\[= \frac{13\pi (4\sqrt{3} - 3)}{8 (16 \times 3 - 9)} = \frac{13\pi (4\sqrt{3} - 3)}{8 (48 - 9)} = \frac{13\pi (4\sqrt{3} - 3)}{8 \times 39}\]

Simplifying using \( 39 = 3 \times 13 \):

\[= \frac{13\pi (4\sqrt{3} - 3)}{8 \times 3 \times 13} = \frac{\pi (4\sqrt{3} - 3)}{24}\]

This matches our computed limit value. Therefore, the second option is correct.

The limit's value is \( \frac{13\pi}{8(4\sqrt{3} + 3)} \).