Animated Solution for Mathematics - Definite Integration: Let ${\lim }_{n\to \infty }\left(\frac{n}{\sqrt{n^4+1}}-\frac{2n}{(n^2+1)\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8n}{(n^2+4)\sqrt{n^4+16}}+\cdots +\frac{n}{\sqrt{n^4+n^4}}-\frac{2n\cdot n^2}{(n^2+n^2)\sqrt{n^4+n^4}}\right)$ be $\frac{\pi }{k}$, using only the principal values of the inverse trigonometric functions. Then $k^2$ is equal to ________

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are peeling back the layers of a complex limit to reveal a hidden, elegant geometry.

When you first look at this expression, it appears as a chaotic, infinite sum of fractions. But remember, in the world of JEE Advanced, chaos is often just order in disguise.

Our journey begins by identifying the heartbeat of this series. We see terms like $\frac{n}{\sqrt{n^4+r^4}}$ and $\frac{2n{r}^2}{(n^2+r^2)\sqrt{n^4+r^4}}$.

By grouping these, we define the general term $T_r$ as:

Look at that numerator! It is the difference of squares. As $n$ approaches infinity, we want to transform this discrete sum into a continuous integral.

To do this, we divide the numerator and denominator by $n^3$, allowing us to express everything in terms of the ratio $\frac{r}{n}$. This is the magic of the Riemann Sum: we are essentially slicing the area under a curve into infinitely thin strips of width $\frac{1}{n}$.

After our algebraic manipulation, the limit $L$ takes the form:

This integral looks intimidating, but let us apply the 'divide by $x^2$' strategy. By dividing both the numerator and the denominator by $x^2$, we prepare the ground for a substitution that will collapse the complexity.

We rewrite the integrand as:

Now, let $t=x+\frac{1}{x}$. The differential $dt=(1-\frac{1}{x^2})dx$ fits perfectly into our numerator.

As $x$ moves from $0$ to $1$, our new variable $t$ moves from $\infty$ to $2$. The integral transforms into:

This is a standard form! Using the integral of the secant inverse, we evaluate this as:

As $t\to \infty$, ${\sec }^{-1}(\infty )=\frac{\pi }{2}$. At the lower bound $t=2$, we have ${\sec }^{-1}(\frac{2}{\sqrt{2}})={\sec }^{-1}(\sqrt{2})=\frac{\pi }{4}$.

Subtracting these, we get:

Given that $L=\frac{\pi }{k}$, we find $k=4\sqrt{2}$. Squaring this value, we arrive at the final result:

What a journey! We started with a daunting limit and, through the power of Riemann sums and clever substitution, reduced it to a simple geometric constant. Remember, the next time you face a problem like this, don't look at the complexity—look for the pattern, trust your tools, and let the math guide you home.