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MathHardAdv 2018 • Paper 2

Evaluation of Series Involving Inverse Tangent Functions and Their Limits

For an arbitrary positive integer nn, let fn:(0,∞)→Rf_n: (0, \infty) \rightarrow \mathbb{R} be defined by fn(x)=∑j=1ntan⁡−1(11+(x+j)(x+j−1))for all x∈(0,∞).f_n(x) = \sum_{j=1}^n \tan^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) \quad \text{for all } x \in (0, \infty). (Assume the inverse trigonometric function tan⁡−1x\tan^{-1} x takes values in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).)

Identify which of the following statements are correct:

Options

A

∑j=15tan⁡2(fj(0))=55\sum_{j=1}^5 \tan^2(f_j(0)) = 55

Correct
B

∑j=110(1+fj′(0))sec⁡2(fj(0))=10\sum_{j=1}^{10} (1 + f'_j(0)) \sec^2(f_j(0)) = 10

Correct
C

For any fixed positive integer nn, lim⁡x→∞tan⁡(fn(x))=1n\lim_{x \to \infty} \tan(f_n(x)) = \frac{1}{n}

D

For any fixed positive integer nn, lim⁡x→∞sec⁡2(fn(x))=1\lim_{x \to \infty} \sec^2(f_n(x)) = 1

Correct

Topics & Concepts

Inverse Trigonometric FunctionsInverse Trigonometric FunctionsLimitsSum of Series

Step-by-Step Solution

To determine which of the given statements are correct, we first simplify the expression for fn(x)f_n(x).

The given function is: fn(x)=∑j=1ntan⁡−1(11+(x+j)(x+j−1))f_n(x) = \sum_{j=1}^n \tan^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right)

We can rewrite the argument of the inverse tangent function using the identity tan⁡−1(a)−tan⁡−1(b)=tan⁡−1(a−b1+ab)\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right): 11+(x+j)(x+j−1)=(x+j)−(x+j−1)1+(x+j)(x+j−1)\frac{1}{1+(x+j)(x+j-1)} = \frac{(x+j) - (x+j-1)}{1 + (x+j)(x+j-1)}

Thus, tan⁡−1(11+(x+j)(x+j−1))=tan⁡−1(x+j)−tan⁡−1(x+j−1)\tan^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) = \tan^{-1}(x+j) - \tan^{-1}(x+j-1)

Substituting this into the sum, we get a telescoping series: fn(x)=∑j=1n[tan⁡−1(x+j)−tan⁡−1(x+j−1)]=tan⁡−1(x+n)−tan⁡−1(x)f_n(x) = \sum_{j=1}^n \left[\tan^{-1}(x+j) - \tan^{-1}(x+j-1)\right] = \tan^{-1}(x+n) - \tan^{-1}(x)

Taking the tangent on both sides: tan⁡(fn(x))=tan⁡(tan⁡−1(x+n)−tan⁡−1(x))=(x+n)−x1+x(x+n)=n1+x2+nx\tan(f_n(x)) = \tan\left(\tan^{-1}(x+n) - \tan^{-1}(x)\right) = \frac{(x+n) - x}{1 + x(x+n)} = \frac{n}{1 + x^2 + nx}

Evaluation of Options:

Option A:

At x=0x = 0: tan⁡(fj(0))=j1+02+0=j\tan(f_j(0)) = \frac{j}{1 + 0^2 + 0} = j

Thus, tan⁡2(fj(0))=j2\tan^2(f_j(0)) = j^2.

The required sum is: ∑j=15tan⁡2(fj(0))=∑j=15j2=12+22+32+42+52=1+4+9+16+25=55\sum_{j=1}^5 \tan^2(f_j(0)) = \sum_{j=1}^5 j^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55

Hence, Option A is correct.

Option B:

Differentiating fn(x)=tan⁡−1(x+n)−tan⁡−1(x)f_n(x) = \tan^{-1}(x+n) - \tan^{-1}(x) with respect to xx: fn′(x)=11+(x+n)2−11+x2f'_n(x) = \frac{1}{1 + (x+n)^2} - \frac{1}{1 + x^2}

At x=0x = 0: fj′(0)=11+j2−11+02=11+j2−1  ⟹  1+fj′(0)=11+j2f'_j(0) = \frac{1}{1 + j^2} - \frac{1}{1 + 0^2} = \frac{1}{1 + j^2} - 1 \implies 1 + f'_j(0) = \frac{1}{1 + j^2}

Since tan⁡(fj(0))=j\tan(f_j(0)) = j, we have: sec⁡2(fj(0))=1+tan⁡2(fj(0))=1+j2\sec^2(f_j(0)) = 1 + \tan^2(f_j(0)) = 1 + j^2

Now, calculating the product: (1+fj′(0))sec⁡2(fj(0))=(11+j2)(1+j2)=1(1 + f'_j(0))\sec^2(f_j(0)) = \left(\frac{1}{1 + j^2}\right)(1 + j^2) = 1

Summing from j=1j = 1 to 1010: ∑j=110(1+fj′(0))sec⁡2(fj(0))=∑j=1101=10\sum_{j=1}^{10} (1 + f'_j(0)) \sec^2(f_j(0)) = \sum_{j=1}^{10} 1 = 10

Hence, Option B is correct.

Option C:

For a fixed positive integer nn: lim⁡x→∞tan⁡(fn(x))=lim⁡x→∞n1+x2+nx=0\lim_{x \to \infty} \tan(f_n(x)) = \lim_{x \to \infty} \frac{n}{1 + x^2 + nx} = 0

Since 0≠1n0 \neq \frac{1}{n}, Option C is incorrect.

Option D:

Using the trigonometric identity sec⁡2(θ)=1+tan⁡2(θ)\sec^2(\theta) = 1 + \tan^2(\theta): lim⁡x→∞sec⁡2(fn(x))=1+lim⁡x→∞tan⁡2(fn(x))=1+02=1\lim_{x \to \infty} \sec^2(f_n(x)) = 1 + \lim_{x \to \infty} \tan^2(f_n(x)) = 1 + 0^2 = 1

Hence, Option D is correct.

Conclusion:

The correct options are A, B, and D.

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