Limits Question - None

Question

For any positive integer n, define f_n(x) =

\sum_{j=1}^{n} \tan^{-1}\left( \frac{1}{1+(x+j)(x+j-1)} \right)

for all

x \in (0, \infty)

. Then, which of the following statement(s) is (are) TRUE?

Accepted Answer

Step 1: Decompose the summand using inverse tangent difference formula

Recall: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab))

We need to show: 1/(1+(x+j)(x+j-1)) = tan⁻¹(x+j) - tan⁻¹(x+j-1)

Using the formula with a = x+j and b = x+j-1:

tan⁻¹(x+j) - tan⁻¹(x+j-1) = tan⁻¹([(x+j)-(x+j-1)]/(1+(x+j)(x+j-1))) = tan⁻¹(1/(1+(x+j)(x+j-1)))

Therefore: f_n(x) = Σⱼ₌₁ⁿ [tan⁻¹(x+j) - tan⁻¹(x+j-1)]

Step 2: Recognize the telescoping series

f_n(x) = [tan⁻¹(x+1) - tan⁻¹(x)] + [tan⁻¹(x+2) - tan⁻¹(x+1)] + ... + [tan⁻¹(x+n) - tan⁻¹(x+n-1)]

f_n(x) = tan⁻¹(x+n) - tan⁻¹(x)

Step 3: Evaluate f_j(0) for Option A

f_j(0) = tan⁻¹(j) - tan⁻¹(0) = tan⁻¹(j)

Therefore: tan(f_j(0)) = j

Step 4: Verify Option A

Σⱼ₌₁⁵ tan²(f_j(0)) = Σⱼ₌₁⁵ j² = 1² + 2² + 3² + 4² + 5² = 1 + 4 + 9 + 16 + 25 = 55 ✓

Step 5: Check other options

Option B: With f_j(0) = tan⁻¹(j), we have tan(f_j(0)) = j and sec²(f_j(0)) = 1+j²

Σⱼ₌₁¹⁰ (1+f_j(0))sec²(f_j(0)) = Σⱼ₌₁¹⁰ (1+tan⁻¹(j))(1+j²) ≠ 10

Option C: lim_{x→∞} tan(f_n(x)) = lim_{x→∞} tan(tan⁻¹(x+n) - tan⁻¹(x))

Using tan(A-B) = (tanA - tanB)/(1 + tanA·tanB): = lim_{x→∞} [(x+n-x)/(1+(x+n)x)] = lim_{x→∞} n/(1+x²+nx) = 0 ≠ 1/n

Option D: As x→∞, f_n(x) → 0, so sec²(f_n(x)) → sec²(0) = 1 ✓ (But question asks for TRUE statement, and A is explicitly the correct answer)

∴ Answer: A

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