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Question: If $\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$ and $S_i = \sum_{k=1}^{\infty} \frac{i}{(36...

If ∑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+S2S_1 + S_2 is equal to

A

π29−12\frac{\pi^2}{9} - \frac{1}{2}

B

π212+12\frac{\pi^2}{12} + \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

We are given Si=∑k=1∞i(36k2−1)iS_i = \sum_{k=1}^{\infty} \frac{i}{(36k^2 - 1)^i}. We need to find S1+S2S_1 + S_2.

Calculating S1S_1: For i=1i=1, S1=∑k=1∞136k2−1S_1 = \sum_{k=1}^{\infty} \frac{1}{36k^2 - 1}. We use the partial fraction decomposition of 136k2−1\frac{1}{36k^2 - 1}. 36k2−1=(6k)2−12=(6k−1)(6k+1)36k^2 - 1 = (6k)^2 - 1^2 = (6k-1)(6k+1). 136k2−1=12(16k−1−16k+1)\frac{1}{36k^2 - 1} = \frac{1}{2} \left( \frac{1}{6k-1} - \frac{1}{6k+1} \right) The sum S1S_1 can be related to a known series summation formula: ∑k=1∞1k2−a2=12a2−πcot⁡(πa)2a\sum_{k=1}^{\infty} \frac{1}{k^2 - a^2} = \frac{1}{2a^2} - \frac{\pi \cot(\pi a)}{2a} Let a=1/6a = 1/6. Then a2=1/36a^2 = 1/36. ∑k=1∞1k2−(1/6)2=∑k=1∞3636k2−1=36S1\sum_{k=1}^{\infty} \frac{1}{k^2 - (1/6)^2} = \sum_{k=1}^{\infty} \frac{36}{36k^2 - 1} = 36 S_1 Using the formula: 36S1=12(1/6)2−πcot⁡(π/6)2(1/6)=12(1/36)−π31/3=18−3π336 S_1 = \frac{1}{2(1/6)^2} - \frac{\pi \cot(\pi/6)}{2(1/6)} = \frac{1}{2(1/36)} - \frac{\pi \sqrt{3}}{1/3} = 18 - 3\pi \sqrt{3} S1=18−3π336=12−π312S_1 = \frac{18 - 3\pi \sqrt{3}}{36} = \frac{1}{2} - \frac{\pi \sqrt{3}}{12}

Calculating S2S_2: For i=2i=2, S2=∑k=1∞2(36k2−1)2=2∑k=1∞1(36k2−1)2S_2 = \sum_{k=1}^{\infty} \frac{2}{(36k^2 - 1)^2} = 2 \sum_{k=1}^{\infty} \frac{1}{(36k^2 - 1)^2}. S2=2∑k=1∞1362(k2−(1/6)2)2=21296∑k=1∞1(k2−(1/6)2)2S_2 = 2 \sum_{k=1}^{\infty} \frac{1}{36^2 (k^2 - (1/6)^2)^2} = \frac{2}{1296} \sum_{k=1}^{\infty} \frac{1}{(k^2 - (1/6)^2)^2} We use the formula: ∑k=1∞1(k2−a2)2=−12a4+πcot⁡(πa)4a3+π2csc⁡2(πa)4a2\sum_{k=1}^{\infty} \frac{1}{(k^2 - a^2)^2} = -\frac{1}{2a^4} + \frac{\pi \cot(\pi a)}{4a^3} + \frac{\pi^2 \csc^2(\pi a)}{4a^2} With a=1/6a = 1/6: a2=1/36a^2 = 1/36, a3=1/216a^3 = 1/216, a4=1/1296a^4 = 1/1296. cot⁡(π/6)=3\cot(\pi/6) = \sqrt{3}, csc⁡(π/6)=2\csc(\pi/6) = 2, csc⁡2(π/6)=4\csc^2(\pi/6) = 4. ∑k=1∞1(k2−(1/6)2)2=−12(1/1296)+π34(1/216)+π2(4)4(1/36)\sum_{k=1}^{\infty} \frac{1}{(k^2 - (1/6)^2)^2} = -\frac{1}{2(1/1296)} + \frac{\pi \sqrt{3}}{4(1/216)} + \frac{\pi^2 (4)}{4(1/36)} =−648+54π3+36π2= -648 + 54\pi \sqrt{3} + 36\pi^2 Now, substitute this back into the expression for S2S_2: S2=21296(−648+54π3+36π2)=1648(−648+54π3+36π2)S_2 = \frac{2}{1296} (-648 + 54\pi \sqrt{3} + 36\pi^2) = \frac{1}{648} (-648 + 54\pi \sqrt{3} + 36\pi^2) S2=−1+54π3648+36π2648=−1+π312+π218S_2 = -1 + \frac{54\pi \sqrt{3}}{648} + \frac{36\pi^2}{648} = -1 + \frac{\pi \sqrt{3}}{12} + \frac{\pi^2}{18}

Calculating S1+S2S_1 + S_2: S1+S2=(12−π312)+(−1+π312+π218)S_1 + S_2 = \left(\frac{1}{2} - \frac{\pi \sqrt{3}}{12}\right) + \left(-1 + \frac{\pi \sqrt{3}}{12} + \frac{\pi^2}{18}\right) S1+S2=12−1−π312+π312+π218S_1 + S_2 = \frac{1}{2} - 1 - \frac{\pi \sqrt{3}}{12} + \frac{\pi \sqrt{3}}{12} + \frac{\pi^2}{18} S1+S2=−12+π218S_1 + S_2 = -\frac{1}{2} + \frac{\pi^2}{18}