Key Concepts and Formulas

• Telescoping Series: A series where intermediate terms cancel out, simplifying the sum. The general form is $\sum_{r=1}^n (a_{r+1} - a_r)$ or $\sum_{r=1}^n (a_r - a_{r+1})$.
• Trigonometric Identity for Arctangent: The identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\frac{x-y}{1+xy}\right)$.
• Limit of Arctangent: $\lim_{x \to \infty} \tan^{-1} x = \frac{\pi}{2}$.
• Limit of a Rational Function: For a rational function where the degree of the numerator and denominator are equal, the limit as $n \to \infty$ is the ratio of the leading coefficients.

Step-by-Step Solution

Step 1: Simplify the general term of the sum. We are given the term $\tan^{-1}\left(\frac{1}{1+r+r^2}\right)$. We can rewrite the argument of the arctangent function to match the form $\frac{x-y}{1+xy}$ from the arctangent subtraction formula. $\frac{1}{1+r+r^2} = \frac{1}{1 + r(r+1)}$ Now, let $x = r+1$ and $y = r$. Then $x-y = (r+1) - r = 1$. So, we can rewrite the term as: $\tan^{-1}\left(\frac{1}{1+r+r^2}\right) = \tan^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}\right)$ Using the arctangent subtraction formula, $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\frac{x-y}{1+xy}\right)$, we get: $\tan^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}\right) = \tan^{-1}(r+1) - \tan^{-1}(r)$

Step 2: Evaluate the sum of the series. The sum is given by $\sum_{r=1}^n \tan^{-1}\left(\frac{1}{1+r+r^2}\right)$. Substituting the result from Step 1, we have a telescoping series: $\sum_{r=1}^n \left(\tan^{-1}(r+1) - \tan^{-1}(r)\right)$ Let's write out the first few terms and the last term to see the cancellation: For $r=1$: $\tan^{-1}(2) - \tan^{-1}(1)$ For $r=2$: $\tan^{-1}(3) - \tan^{-1}(2)$ For $r=3$: $\tan^{-1}(4) - \tan^{-1}(3)$ ... For $r=n$: $\tan^{-1}(n+1) - \tan^{-1}(n)$

When we sum these terms, the intermediate terms cancel out: $(\tan^{-1}(2) - \tan^{-1}(1)) + (\tan^{-1}(3) - \tan^{-1}(2)) + (\tan^{-1}(4) - \tan^{-1}(3)) + \dots + (\tan^{-1}(n+1) - \tan^{-1}(n))$ The sum simplifies to: $\tan^{-1}(n+1) - \tan^{-1}(1)$

Step 3: Apply the arctangent subtraction formula to the simplified sum. We have the sum as $\tan^{-1}(n+1) - \tan^{-1}(1)$. We can use the arctangent subtraction formula again, with $x = n+1$ and $y = 1$. $\tan^{-1}(n+1) - \tan^{-1}(1) = \tan^{-1}\left(\frac{(n+1)-1}{1+(n+1)(1)}\right)$ $= \tan^{-1}\left(\frac{n}{1+n+1}\right)$ $= \tan^{-1}\left(\frac{n}{n+2}\right)$

Step 4: Evaluate the limit of the tangent of the sum. We need to find the limit of $\tan \left(\sum_{r=1}^n \tan^{-1}\left(\frac{1}{1+r+r^2}\right)\right)$ as $n \to \infty$. From Step 3, the sum is equal to $\tan^{-1}\left(\frac{n}{n+2}\right)$. So, we need to evaluate: $\mathop {\lim }\limits_{n \to \infty } \tan \left( \tan^{-1}\left(\frac{n}{n+2}\right) \right)$ Since $\tan(\tan^{-1}(x)) = x$ for all real $x$, we have: $\tan \left( \tan^{-1}\left(\frac{n}{n+2}\right) \right) = \frac{n}{n+2}$ Now, we evaluate the limit of this expression as $n \to \infty$: $\mathop {\lim }\limits_{n \to \infty } \frac{n}{n+2}$ To evaluate this limit, we can divide the numerator and the denominator by the highest power of $n$ in the denominator, which is $n$: $\mathop {\lim }\limits_{n \to \infty } \frac{n/n}{(n+2)/n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{1 + 2/n}$ As $n \to \infty$, $\frac{2}{n} \to 0$. Therefore, the limit becomes: $\frac{1}{1 + 0} = 1$

Common Mistakes & Tips

• Incorrectly applying the arctangent formula: Ensure the form $\frac{x-y}{1+xy}$ is correctly matched to the given term. Mistakes in identifying $x$ and $y$ will lead to an incorrect sum.
• Errors in telescoping sum cancellation: Carefully write out the terms to ensure proper cancellation. Missing or incorrectly cancelling terms will result in a wrong final sum.
• Confusing $\lim_{n \to \infty} \tan^{-1}(expression)$ with $\tan(\lim_{n \to \infty} expression)$: While $\tan(\tan^{-1}(x)) = x$, the limit of the tangent of an arctangent is not always straightforward if the argument of the arctangent does not simplify nicely. In this case, the simplification was possible.

Summary

The problem requires evaluating a limit involving a sum of arctangent terms. The key step is to recognize that the general term of the sum, $\tan^{-1}\left(\frac{1}{1+r+r^2}\right)$, can be rewritten as a difference of two arctangent terms, $\tan^{-1}(r+1) - \tan^{-1}(r)$, using the arctangent subtraction formula. This transforms the sum into a telescoping series, which simplifies to $\tan^{-1}(n+1) - \tan^{-1}(1)$. Further application of the arctangent subtraction formula yields $\tan^{-1}\left(\frac{n}{n+2}\right)$. Finally, we take the limit of the tangent of this expression as $n \to \infty$. Since $\tan(\tan^{-1}(x)) = x$, the problem reduces to finding the limit of $\frac{n}{n+2}$ as $n \to \infty$, which is 1.

The final answer is \boxed{1}.