For any positive integer n, define fn : (0, ∞) → R as fn(x) = Σ(j=1 to n) tan⁻¹(1/(1+(x+j)(x+j-1)))

Q: Wait, why can't we just plug in $x=0$ to evaluate options A and B?

That is a classic trap! The problem explicitly defines the domain of $f_n(x)$ as $(0, \infty)$. This means $x=0$ is strictly excluded from the domain. Always check the domain before substituting values; in this case, $f_j(0)$ is undefined, making those options invalid.

Q: How do I know when to use the telescoping series trick?

Whenever you see a summation involving inverse trigonometric functions where the argument is a fraction of the form $\frac{1}{1+AB}$, it is almost always a hint to use the $\tan^{-1}A - \tan^{-1}B$ identity. If the difference between $A$ and $B$ is constant (or related to the numerator), you are likely looking at a telescoping series.

Q: Is the limit of $\tan^{-1}(x+n) - \tan^{-1}(x)$ always zero as $x \to \infty$?

Yes, it is. As $x$ approaches infinity, both $(x+n)$ and $x$ approach infinity. Since $\lim_{t\to\infty} \tan^{-1}(t) = \frac{\pi}{2}$, the expression becomes $\frac{\pi}{2} - \frac{\pi}{2}$, which is exactly $0$. This is a powerful result that simplifies the limit analysis for the remaining options.

Question

For any positive integer $n$, define $f_n : (0, \infty) 	o R$ as $f_n(x) = \sum_{j=1}^{n} 	an^{-1}\left(\frac{1}{1+(x+j)(x+ j -1)}ight)$ for all $x \in (0,\infty)$. Here, the inverse trigonometric function $	an^{-1}x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$. Then, which of the following statement(s) is (are) TRUE ?

Options:

(A) $\sum_{j=1}^{n} 	an^2 (f_j (0)) = 55$

(B) $\sum_{j=1}^{n} (1+f_j(0))\sec^2 (f_j (0)) = 10$

(C) For any fixed positive integer $n$, $\lim_{x	o\infty} 	an (f_n(x)) = \frac{1}{n}$

(D) For any fixed positive integer $n$, $\lim_{x	o\infty} \sec^2 (f_n(x)) = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, JEE aspirants! Today, we are going to dissect a problem that looks intimidating at first glance but reveals a beautiful, elegant structure once we peel back the layers.

We are dealing with a function defined as:
$$f_n(x) = \sum_{j=1}^{n} 	an^{-1}\left(\frac{1}{1+(x+j)(x+ j - 1)}ight)$$

At first, this summation might seem like a nightmare to compute. But in the world of JEE Advanced, whenever you see a complex expression inside a summation, there is almost always a hidden pattern waiting to be discovered.

### The Anatomy of the Term

Let us focus on the general term of our series, $T_j$:
$$T_j = 	an^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}ight)$$

The denominator is the key. We have $1$ plus the product of two terms: $(x+j)$ and $(x+j-1)$.

Now, consider the difference between these two factors:
$$(x+j) - (x+j-1) = 1$$

This is not a coincidence; it is a deliberate design. We can rewrite the numerator, which is $1$, as the difference of these two factors. So, our term becomes:
$$T_j = 	an^{-1}\left(\frac{(x+j) - (x+j-1)}{1+(x+j)(x+j-1)}ight)$$

### The Identity of Elegance

This structure is a perfect match for the inverse trigonometric identity:
$$	an^{-1}\left(\frac{A-B}{1+AB}ight) = 	an^{-1}A - 	an^{-1}B$$

By identifying $A = x+j$ and $B = x+j-1$, we can transform our complex term into a simple difference:
$$T_j = 	an^{-1}(x+j) - 	an^{-1}(x+j-1)$$

This is the "Aha!" moment. We have turned a product-based fraction into a simple subtraction.

### The Telescoping Magic

Now, let us expand the summation $f_n(x) = \sum_{j=1}^{n} (	an^{-1}(x+j) - 	an^{-1}(x+j-1))$. When we write out the terms, we get:

For $j=1$: $	an^{-1}(x+1) - 	an^{-1}(x)$
For $j=2$: $	an^{-1}(x+2) - 	an^{-1}(x+1)$
...
For $j=n$: $	an^{-1}(x+n) - 	an^{-1}(x+n-1)$

Notice the beautiful chain reaction: the positive part of the first term cancels the negative part of the second, the positive part of the second cancels the negative part of the third, and this continues until only the very last positive term and the very first negative term remain.

Thus, the simplified function is:
$$f_n(x) = 	an^{-1}(x+n) - 	an^{-1}(x)$$

### The Domain Trap

Before we rush to evaluate the options, we must pause. Options A and B ask us to calculate values at $x=0$.

However, the problem defines the domain as $x \in (0, \infty)$. Since $x=0$ is not in the domain, any expression involving $f_j(0)$ is mathematically invalid. Do not fall into the trap of blindly substituting values!

### The Limit at Infinity

Finally, for Options C and D, we need the limit as $x 	o \infty$. We have:
$$\lim_{x	o\infty} f_n(x) = \lim_{x	o\infty} (	an^{-1}(x+n) - 	an^{-1}(x))$$

As $x$ grows without bound, both $	an^{-1}(x+n)$ and $	an^{-1}(x)$ approach $\frac{\pi}{2}$. Therefore, the limit is:
$$\frac{\pi}{2} - \frac{\pi}{2} = 0$$

Now, checking the options: Option C claims the limit of $	an(f_n(x))$ is $\frac{1}{n}$, but $	an(0) = 0$, so this is false.

Option D claims the limit of $\sec^2(f_n(x))$ is $1$, and since $\sec^2(0) = 1$, this is **perfectly true**. Through careful observation and the power of telescoping, we have arrived at the correct conclusion.

Select Answer:

Visualized Solution

General Term of $f_{n} (x)$

Manipulating the Numerator

Applying Inverse Trig Identity

Expanding the Telescoping Series

Canceling Intermediate Terms

Checking Options A and B

Evaluating Limit at Infinity

Checking Option C

Checking Option D

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Anatomy of the Term

The Identity of Elegance

The Telescoping Magic

The Domain Trap

The Limit at Infinity

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