Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of ${\lim }_{n\to \infty }6\tan \left({\sum }_{r=1}^n{\tan }^{-1}\left(\frac{1}{r^2+3r+3}\right)\right)$ is equal to

Imagine standing before a mountain of a problem. You see a limit, a summation, and a nested inverse tangent function. It looks like a chaotic mess, but in the world of JEE Advanced, complexity is often just a mask for elegance.

Let's peel back that mask together.

Our journey begins with the general term:

When you see an inverse tangent in a summation, your brain should immediately scream: "Telescoping Series!" The goal is to transform this single term into a difference of two terms, $f(r+1)-f(r)$.

We need to use the identity:

To do this, we need the denominator to look like $1+xy$. Look at $r^2+3r+3$. If we split the $3$ into $1+2$, we get $1+(r^2+3r+2)$.

Factoring the quadratic $r^2+3r+2$ gives us $(r+1)(r+2)$. Now, the denominator is $1+(r+1)(r+2)$. This is perfect!

Now, look at the numerator. We have a $1$. Does it match $x-y$? If $x=r+2$ and $y=r+1$, then $x-y=(r+2)-(r+1)=1$. It matches perfectly!

So, our general term $T_r$ becomes:

Now, let's expand the sum $S_n={\sum }_{r=1}^n[{\tan }^{-1}(r+2)-{\tan }^{-1}(r+1)]$. When we write out the terms, we see:

Notice how the positive ${\tan }^{-1}3$ cancels with the negative ${\tan }^{-1}3$? This domino effect continues, leaving us with only:

We are almost there. The problem asks for the limit of $6\tan (S_n)$. We need to find:

Using the tangent subtraction formula, this simplifies to:

Finally, we take the limit as $n\to \infty$:

Dividing by $n$, we get $6\cdot \frac{1}{2+5/n}$. As $n$ approaches infinity, $5/n$ vanishes, leaving us with:

You did it! You navigated the complexity and found the simple, elegant truth hidden within.