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

Answer: (A)

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

Answer 2

article amsmath

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:

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

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