Animated Solution for Mathematics - Definite Integration: Let $S_n={\sum }_{k=1}^n\frac{n}{n^2+kn+k^2}$ and $T_n={\sum }_{k=0}^{n-1}\frac{n}{n^2+kn+k^2}$ for $n=1,2,3,\dots$. Then,

A Masterclass in Riemann Sums

My dear students, welcome to a journey through one of the most beautiful intersections in mathematics: the bridge between the discrete world of summations and the continuous world of calculus. Today, we are not just solving a problem; we are learning to see the hidden geometry behind algebraic expressions.

We are given two sums, $S_n$ and $T_n$, and asked to compare them to a specific constant, $\frac{\pi }{3\sqrt{3}}$. At first glance, these sums look like intimidating algebraic monsters.

But I want you to take a deep breath. In the world of JEE Advanced, when you see a sum where the number of terms $n$ goes to infinity, your mind should immediately scream: 'Riemann Sum!'

The Transformation

Let us look at the general term of our summation: $\frac{n}{n^2+kn+k^2}$. It looks messy, doesn't it? But let us apply the 'JEE lens' to it.

We want to transform this into a function of $x=\frac{k}{n}$. To do this, we divide both the numerator and the denominator by $n^2$. Watch the magic happen:

Suddenly, the chaos organizes itself. We have a factor of $\frac{1}{n}$, which represents our differential width $dx$, and a function $f(x)=\frac{1}{1+x+x^2}$ where $x=\frac{k}{n}$. We have successfully translated a discrete sum into the language of calculus!

The Geometry of Rectangles

Now, imagine you are standing on a graph. You have the curve $f(x)=\frac{1}{1+x+x^2}$ plotted from $x=0$ to $x=1$. A Riemann sum is simply a way of approximating the area under this curve using rectangles.

For $S_n$, the sum starts at $k=1$ and ends at $k=n$. This means we are evaluating the function at the right endpoints of our sub-intervals.

Because our function $f(x)$ is strictly decreasing (as we can verify by its derivative $f^{\prime }(x)=-\frac{1+2x}{(1+x+x^2{)}^2}$), the right endpoint of any interval is the lowest point of the curve in that interval. Therefore, the rectangles for $S_n$ sit entirely under the curve. This is our 'Right Riemann Sum', and it must be less than the true area under the curve.

For $T_n$, the sum starts at $k=0$ and ends at $k=n-1$. Here, we are evaluating the function at the left endpoints. Since the function is decreasing, the left endpoint is the highest point of the curve in that interval.

Thus, the rectangles for $T_n$ stick out above the curve. This is our 'Left Riemann Sum', and it must be greater than the true area under the curve.

The Benchmark Integral

To prove these inequalities, we need the 'true area'—the definite integral. We must evaluate:

This is a classic integral. We complete the square in the denominator: $1+x+x^2=(x+\frac{1}{2}{)}^2+\frac{3}{4}=(x+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2$. Now, the integral becomes:

Using the standard integral $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}(\frac{x}{a})$, we get:

We know that ${\tan }^{-1}(\sqrt{3})=\frac{\pi }{3}$ and ${\tan }^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi }{6}$. Subtracting these gives us $\frac{\pi }{6}$. Multiplying by $\frac{2}{\sqrt{3}}$, we arrive at the beautiful result: $I=\frac{\pi }{3\sqrt{3}}$.

The Final Victory

We have our benchmark! Because $f(x)$ is decreasing, our Right Riemann Sum $S_n$ is an underestimate, so $S_n<\frac{\pi }{3\sqrt{3}}$. Our Left Riemann Sum $T_n$ is an overestimate, so $T_n>\frac{\pi }{3\sqrt{3}}$.

Do you see the elegance? We didn't just calculate numbers; we visualized the behavior of a function and used the power of calculus to bound our sums. This is the essence of JEE Advanced physics and mathematics. Keep this geometric intuition close to your heart, and no problem will ever be too difficult for you. You have mastered the Riemann sum!