If Uₙ = ( 1 + 1/n² )( 1 + 2²/n² )².....( 1 + n²/n² )ⁿ, then lim _(n → ∞) (Uₙ)^((- 4)/n²) is equal to :

Solution

Maths sir · 4 steps

Take logs: the product becomes a sum, and that sum is a Riemann sum for an integral.

1. Take the log. Write Uₙ = Πᵣ₌₁ⁿ (1 + r²/n²)ʳ and let L be the required limit. Then ln L = lim(n → ∞) -4/n² Σᵣ₌₁ⁿ r ln(1 + r²/n²).
2. Spot the Riemann sum. Split r/n² as 1/n · r/n and put x = r/n. As n → ∞, 1/nΣ becomes ∫₀¹ dx.

   ln L = -4∫₀¹ x ln(1 + x²) dx

3. Evaluate the integral. Put t = 1 + x², so x dx = 1/2dt and the limits become 1 to 2. The integral is 1/2[t ln t - t]₁² = 1/2(2 ln 2 - 1) = ln 2 - 1/2.
4. So. ln L = -4(ln 2 - 1/2) = 2 - 4 ln 2, so L = e² · e⁻⁴ˡⁿ² = e²/16, which is option (a).

The idea to keep. A limit of a long product raised to a power of n: take logs, write the sum as 1/nΣ f(r/n), and it becomes ∫₀¹ f(x) dx.

Answer from the NTA final answer key. Solution by PrepWiser.