Animated Solution for Mathematics - Definite Integration: The value of ${\lim }_{n\to \infty }{\sum }_{k=1}^n\frac{n^3}{(n^2+k^2)(n^2+3k^2)}$ is:

Bridging the Discrete and Continuous

Welcome, fellow traveler on the path to JEE mastery. Today, we stand before a problem that might seem like a daunting mountain of algebra, but it is actually a beautifully crafted staircase.

We are looking at the limit of a sum: ${\lim }_{n\to \infty }{\sum }_{k=1}^n\frac{n^3}{(n^2+k^2)(n^2+3k^2)}$.

When you see a limit of a sum as $n$ approaches infinity, your mathematical intuition should immediately scream, "Riemann Sum!" This is the bridge between the discrete world of individual terms and the continuous world of calculus.

We are essentially summing up the areas of infinitely many, infinitely thin rectangles to find the total area under a curve. Our grand strategy is to transform this discrete summation into a smooth, continuous definite integral.

The $n^4$ Trick

To make this transformation, we need to mold our expression into a specific format. We are hunting for terms that look like $\frac{k}{n}$ and a standalone $\frac{1}{n}$.

Because $\frac{k}{n}$ will become our variable $x$, and $\frac{1}{n}$ will become our infinitesimally small width, $dx$. Look at our expression: $\frac{n^3}{(n^2+k^2)(n^2+3k^2)}$.

If we divide both the numerator and the denominator by $n^4$, watch how the magic happens:

Suddenly, the structure is revealed! We have our $\frac{1}{n}$ factor, and the rest of the expression is a function of $\frac{k}{n}$. This is the "algebraic surgery" that makes the problem solvable.

Now, visualize this: as $n$ becomes infinitely large, those discrete rectangles we just formed become infinitely thin. The sum transforms into an integral!

We replace $\frac{k}{n}$ with $x$, and $\frac{1}{n}$ with $dx$. For the limits, when $k=1$, $x$ approaches $0$. When $k=n$, $x$ approaches $1$.

Our complex summation has elegantly collapsed into the integral:

The Art of Simplification

We now have a rational function to integrate. The denominator is a product of two quadratic terms, which is a textbook case for partial fractions.

To make our lives easier and avoid messy algebra, let's temporarily substitute $x^2$ with a dummy variable $y$. This turns our expression into $\frac{1}{(1+y)(1+3y)}$.

We split this into two simpler fractions:

By solving for $A$ and $B$, we find $A=-\frac{1}{2}$ and $B=\frac{3}{2}$. Substituting $x^2$ back in, our integrand becomes:

Integration and Rationalization

Now, we apply standard integration formulas. The integral of $\frac{1}{1+x^2}$ is ${\tan }^{-1}x$, and the integral of $\frac{1}{1+3x^2}$ involves $\frac{1}{\sqrt{3}}{\tan }^{-1}(\sqrt{3}x)$.

Evaluating this from $0$ to $1$ gives us:

Note: The final result simplifies to $\frac{\pi }{24}$.

Remember, the key is always pattern recognition. Whenever you face an infinite sum, look for that $\frac{1}{n}$ and form a function of $\frac{k}{n}$. That is your golden ticket.