Tackling a Tricky JEE Problem

Introduction

• Topic: Solving a challenging math problem from the JEE exam.

• Why It's Important: JEE is known to be one of the toughest exams globally, particularly in math.

• The problem deals with a sum denoted as SN defined as:

  SN = ∑ from k=1 to n of n / (n^2 + kn + k^2)
  

  for n in natural numbers (1, 2, 3, ...).

• Objective: Determine whether S(n) is less than or greater than π / (3√3).

Analyzing the Problem

Key Concepts and Inequalities

• Analyzing the Bound: The maximum value that SN could take is its limit as n approaches infinity, denoted by S(∞).
• Strategy:
  1. Factor out an n to simplify the expression.
  2. Change the summation to a form involving fractions more approachable for evaluation.
  3. Identify connections to Riemann sums.

Transforming the Sum

• Rewrite SN as:

  limit as n -> ∞ of ∑ from k=1 to n of (1 / (n * (1 + k/n + (k/n)^2)))
  

Connecting to Riemann Sums

• A Riemann Sum transforms the sum into an integral form.

• General Riemann Sum:

  ∫ from a to b of f(x) dx ≈ limit as n -> ∞ of ∑ from k=1 to n of f(x_k) * Δx
  

• Identify Δx = 1/n and x_k = k/n for the function f(x) = 1 / (1 + x + x^2).

• Bounds: From 0 to 1 when a=0 and b=1.

• Transform the sum into an integral:

  S(∞) = ∫ from 0 to 1 of 1 / (1 + x + x^2) dx
  

Solving the Integral

• Complete the square for the integral’s denominator:

  1 / (x^2 + x + 1) = 1 / ((x + 1/2)^2 + 3/4)
  

• Recognize this as an arctan integral:

  ∫ 1 / (a^2 + u^2) du = (1/a) * arctan(u/a)
  

• Apply the formula to solve the integral, adjusting for constants:

  ∫ from 0 to 1 of 1 / ((x + 1/2)^2 + 3/4) dx = 2 / √3 * [arctan((2x + 1) / √3)] from 0 to 1
  

Evaluating the Integral

• Evaluate at the bounds x=0 and x=1:

  = 2 / √3 * [arctan(3 / √3) - arctan(1 / √3)]
  = 2 / √3 * [π/3 - π/6]
  = π / (3√3)
  

• Therefore, S(n) < S(∞) and S(∞) = π / (3√3).

Conclusion

• Final Result: SN < π / (3√3).
• Challenging yet rewarding problem involving Riemann sums and integral calculus.
• Challenge: Solve a similar inequality for another sum T(n) defined with a different range.

Mathematical Formulas Used

• Riemann Sum:

  ∫ from a to b of f(x) dx ≈ limit as n -> ∞ of ∑ from k=1 to n of f(x_k) * Δx
  

• Arctan Integral:

  ∫ 1 / (a^2 + u^2) du = (1/a) * arctan(u/a)
  

Reminder

• Ensure you are comfortable with Riemann sums and integral calculus to tackle similar challenging problems.

Challenge: Find similar inequalities for T(n) starting from 0 to n-1.