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

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To solve the limit

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

we start by rewriting it as a Riemann sum.

Step 1: Rewrite as a Riemann Sum

Rewrite each term in the sum by dividing both terms inside the denominator by \(n^2\):

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

As \(n\to\infty\), the term \(\frac{k}{n}\) behaves like a continuous variable \(x\) on the interval \([0,1]\). Thus, the sum can be approximated by the integral:

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

Step 2: Simplify the Integral

We now need to evaluate:

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

Step 3: Use Partial Fraction Decomposition

To simplify this integral, we can use partial fraction decomposition. We assume that:

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

Multiplying both sides by \((1+x^2)(1+3x^2)\) gives:

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

Expanding and combining terms, we get:

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

Solving this system:

From \(A+B=1\), we get \(B=1-A\).

Substitute \(B=1-A\) in \(3A+B=0\):

\[3A+(1-A)=0 \Rightarrow 2A=-1 \Rightarrow A=\frac{-1}{3}\]

Substitute \(A=\frac{-1}{3}\) into \(B=1-A\):

\[B=1-\frac{-1}{3}=\frac{1}{3}\]

Thus, we have:

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

Step 4: Rewrite the Integral

Now the integral becomes:

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

Step 5: Evaluate Each Integral Separately

1. First integral:
2. Second integral: Use the substitution \(u=\sqrt{3}x\), then \(du=\sqrt{3}dx\), or \(dx=\frac{du}{\sqrt{3}}\).

Step 6: Combine the Results

Now, substitute back into the integral:

\[\int_0^1 \frac{1}{(1+x^2)(1+3x^2)} \,dx = \frac{1}{3} \times \frac{\pi}{4} - \frac{1}{3} \times \frac{\pi}{3\sqrt{3}}\]

\[= \frac{\pi}{12} - \frac{\pi}{9\sqrt{3}} = \frac{13\pi}{8(4\sqrt{3}+3)}\]