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

Imagine you are standing on the edge of a vast mathematical landscape. You are presented with a problem that looks like a tangled mess:

At first glance, it seems impossible to sum an infinite number of inverse tangent functions. But here is the secret: mathematics is rarely about brute force; it is about finding the hidden structure. This problem is a classic example of a telescoping series, a beautiful phenomenon where a long, complex chain of terms collapses into just a few survivors.

Deconstructing the General Term

Our journey begins by isolating the general term, $T_r={\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)$. When you see an inverse tangent function in a summation, your intuition should immediately scream, "Can I turn this into a difference of two terms?"

We want to use the identity:

Look at the denominator: $1+r+r^2$. If we factor out $r$ from the last two terms, we get $1+r(r+1)$.

This is the "Aha!" moment. It perfectly matches the $1+xy$ structure of our identity, where $x=r+1$ and $y=r$.

Since $(r+1)-r=1$, and our numerator is already $1$, we can rewrite the entire term as:

The Telescoping Collapse

With our identity applied, the general term becomes $T_r={\tan }^{-1}(r+1)-{\tan }^{-1}(r)$. Now, let's watch the magic unfold as we expand the sum $S_n={\sum }_{r=1}^n{T}_r$.

For $r=1$, we have ${\tan }^{-1}(2)-{\tan }^{-1}(1)$.

For $r=2$, we have ${\tan }^{-1}(3)-{\tan }^{-1}(2)$.

For $r=3$, we have ${\tan }^{-1}(4)-{\tan }^{-1}(3)$.

As you write these out, you can see the pattern. The positive ${\tan }^{-1}(2)$ from the first term cancels the negative ${\tan }^{-1}(2)$ from the second term. The positive ${\tan }^{-1}(3)$ from the second term cancels the negative ${\tan }^{-1}(3)$ from the third.

This chain reaction, the telescoping effect, slashes through all intermediate terms. It leaves only the negative part of the first term and the positive part of the last term:

The Limit and the Final Act

We are almost there. We need to find the limit as $n\to \infty$. As $n$ grows, $n+1$ approaches infinity, and we know that ${\lim }_{x\to \infty }{\tan }^{-1}(x)=\frac{\pi }{2}$.

Meanwhile, ${\tan }^{-1}(1)$ is a constant, $\frac{\pi }{4}$. Thus, the limit of our sum is:

Finally, we evaluate the tangent of this limit value:

This problem is a masterclass in pattern recognition. It teaches us that even the most intimidating expressions can be tamed if we look for the underlying structure. The final answer is 1.