Question

$\tan^{-1} \left( \frac{1}{1 + 1 \cdot 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \cdot 3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n \cdot (n+1)} \right) =$

• A. $\tan^{-1} \left( \frac{n}{n+2} \right)$
• B. $\tan^{-1} \left( \frac{n+1}{n} \right)$
• C. $\tan^{-1} \left( \frac{n}{n+1} \right)$
• D. $\tan^{-1} \left( \frac{n+2}{n} \right)$

Hint:

For series summation involving $\tan^{-1}$, always try to express the argument in the form $\frac{x-y}{1+xy}$. This allows you to split the term into $\tan^{-1}x - \tan^{-1}y$, which almost always leads to a telescoping series where most terms neatly cancel out.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires evaluating the sum of a series involving inverse trigonometric functions. The general strategy is to express each term as a difference of two inverse tangents to create a telescoping series, where intermediate terms cancel out.

Step 2: Key Formula or Approach:
Use the inverse tangent difference formula: \[ \tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \] We need to rewrite the $k$-th term, $T_k = \tan^{-1} \left( \frac{1}{1 + k(k+1)} \right)$, in a form that matches the right side of this identity.

Step 3: Detailed Explanation:
Let's analyze the general $k$-th term of the series: \[ T_k = \tan^{-1} \left( \frac{1}{1 + k(k+1)} \right) \] We can cleverly rewrite the numerator $1$ as $(k+1) - k$. \[ T_k = \tan^{-1} \left( \frac{(k+1) - k}{1 + k \cdot (k+1)} \right) \] Comparing this with the formula $\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}x - \tan^{-1}y$, we can see that $x = k+1$ and $y = k$. Therefore, we can write $T_k$ as a difference: \[ T_k = \tan^{-1}(k+1) - \tan^{-1}(k) \] Now, we need to find the sum of the first $n$ terms, $S_n = \sum_{k=1}^n T_k$: \[ S_n = \sum_{k=1}^n [\tan^{-1}(k+1) - \tan^{-1}(k)] \] Writing out the terms to see the pattern: For $k=1$: $\tan^{-1}(2) - \tan^{-1}(1)$
For $k=2$: $\tan^{-1}(3) - \tan^{-1}(2)$
For $k=3$: $\tan^{-1}(4) - \tan^{-1}(3)$
$\dots$
For $k=n$: $\tan^{-1}(n+1) - \tan^{-1}(n)$
Adding all these terms together, we see that almost all terms cancel out (this is a telescoping sum): \[ S_n = [\tan^{-1}(2) - \tan^{-1}(1)] + [\tan^{-1}(3) - \tan^{-1}(2)] + \dots + [\tan^{-1}(n+1) - \tan^{-1}(n)] \] \[ S_n = \tan^{-1}(n+1) - \tan^{-1}(1) \] We know that $\tan^{-1}(1) = \frac{\pi}{4}$, but it's more useful to apply the difference formula again to combine these two terms into a single inverse tangent expression: \[ S_n = \tan^{-1}(n+1) - \tan^{-1}(1) = \tan^{-1} \left( \frac{(n+1) - 1}{1 + (n+1) \cdot 1} \right) \] \[ S_n = \tan^{-1} \left( \frac{n}{1 + n + 1} \right) \] \[ S_n = \tan^{-1} \left( \frac{n}{n+2} \right) \]
Step 4: Final Answer:
The sum of the series is $\tan^{-1} \left( \frac{n}{n+2} \right)$.