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Question: Let $\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$ and $S_i = \sum_{k=1}^{\infty} \frac{i}{(3...

Let ∑k=1∞1k2=π26\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} and Si=∑k=1∞i(36k2−1)iS_i = \sum_{k=1}^{\infty} \frac{i}{(36k^2 - 1)^i}, then

S1+S2=S_1 + S_2 =

A

π212+12\frac{\pi^2}{12} + \frac{1}{2}

B

π29−12\frac{\pi^2}{9} - \frac{1}{2}

C

π215+12\frac{\pi^2}{15} + \frac{1}{2}

D

π218−12\frac{\pi^2}{18} - \frac{1}{2}

Answer

π218−12\frac{\pi^2}{18} - \frac{1}{2}

Explanation

Solution

Solution Explanation:

  1. Write
      S₁ = ∑ₖ₌₁∞ 1/(36k² – 1) = ∑ₖ₌₁∞ 1/[(6k–1)(6k+1)].
      Using partial fractions,
        1/(6k–1)(6k+1) = ½[1/(6k–1) – 1/(6k+1)].

  2. Express S₁ in a form involving ∑ₖ₌₁∞ 1/(k² – 1/36) by noting
      36k² – 1 = 36[k² – (1/6)²].
      Thus, S₁ = 1/36 ∑ₖ₌₁∞ 1/(k² – (1/6)²).

  3. Use the known formula (valid for a non‐integer a):
      ∑ₖ₌₁∞ 1/(k² – a²) = 1/(2a²) – (π/(2a))cot(πa).
      For a = 1/6, one obtains
        S₁ = 1/36 [ 18 – 3π cot(π/6) ] = (6 – π√3)/12
        (since cot(π/6)=√3).

  4. Similarly, express
      S₂ = ∑ₖ₌₁∞ 2/(36k² – 1)² = 2/36² ∑ₖ₌₁∞ 1/(k² – (1/6)²)².

  5. Use the formula
      ∑ₖ₌₁∞ 1/(k² – a²)² = [π/(4a³)]cot(πa) + [π²/(4a²)]csc²(πa) – 1/(2a⁴).
      For a = 1/6, compute the terms noting:
        (1/6)³ = 1/216, (1/6)⁴ = 1/1296, cot(π/6)=√3, csc(π/6)=2, csc²(π/6)=4.
      This yields
        ∑ₖ₌₁∞ 1/(k² – (1/6)²)² = 54π√3 + 36π² – 648.
      Thus,
        S₂ = 2/(1296) (54π√3 + 36π² – 648) = (6π√3 + 4π² – 72)/72.

  6. Combine S₁ and S₂ with common denominator (convert S₁ to denominator 72):
      S₁ = (6 – π√3)/12 = (36 – 6π√3)/72.
      Then,
        S₁ + S₂ = [ (36 – 6π√3) + (6π√3 + 4π² – 72) ]/72
             = (4π² – 36)/72 = (π² – 9)/18.
      This can be written as
        π²/18 – 1/2.

Answer:
Option (d):  π218−12\displaystyle \frac{\pi^2}{18}-\frac{1}{2}