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Question: given: $\sum_{R=1}^{\infty} \frac{1}{R^2} = \frac{\pi^2}{6}.$ $S_i = \sum_{k=1}^{\infty} \frac{1}{(...

given: ∑R=1∞1R2=π26.\sum_{R=1}^{\infty} \frac{1}{R^2} = \frac{\pi^2}{6}.

Si=∑k=1∞1(36k2−1)∗iS_i = \sum_{k=1}^{\infty} \frac{1}{(36k^2-1)} * i

find S1+S2S_1 + S_2

Answer

32−π34\frac{3}{2} - \frac{\pi \sqrt{3}}{4}

Explanation

Solution

Let S=∑k=1∞136k2−1S = \sum_{k=1}^{\infty} \frac{1}{36k^2-1}. The expression SiS_i is given by Si=i⋅SS_i = i \cdot S. We need to find S1+S2=1⋅S+2⋅S=3SS_1 + S_2 = 1 \cdot S + 2 \cdot S = 3S.

To evaluate SS, we use the partial fraction decomposition of the term 136k2−1\frac{1}{36k^2-1}. Using the difference of squares formula, 36k2−1=(6k)2−12=(6k−1)(6k+1)36k^2-1 = (6k)^2 - 1^2 = (6k-1)(6k+1). We can write: 136k2−1=1(6k−1)(6k+1)\frac{1}{36k^2-1} = \frac{1}{(6k-1)(6k+1)} Using partial fractions: 1(6k−1)(6k+1)=A6k−1+B6k+1\frac{1}{(6k-1)(6k+1)} = \frac{A}{6k-1} + \frac{B}{6k+1} Multiplying by (6k−1)(6k+1)(6k-1)(6k+1): 1=A(6k+1)+B(6k−1)1 = A(6k+1) + B(6k-1) Setting 6k−1=0  ⟹  k=1/66k-1=0 \implies k=1/6: 1=A(1+1)  ⟹  2A=1  ⟹  A=1/21 = A(1+1) \implies 2A = 1 \implies A = 1/2. Setting 6k+1=0  ⟹  k=−1/66k+1=0 \implies k=-1/6: 1=B(−1−1)  ⟹  −2B=1  ⟹  B=−1/21 = B(-1-1) \implies -2B = 1 \implies B = -1/2. So, 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) Now, we can evaluate the sum SS: S=∑k=1∞12(16k−1−16k+1)=12∑k=1∞(16k−1−16k+1)S = \sum_{k=1}^{\infty} \frac{1}{2} \left( \frac{1}{6k-1} - \frac{1}{6k+1} \right) = \frac{1}{2} \sum_{k=1}^{\infty} \left( \frac{1}{6k-1} - \frac{1}{6k+1} \right) This sum can be evaluated using the Mittag-Leffler expansion for the cotangent function, or by relating it to a known series summation formula. A standard result for sums of the form ∑n=1∞1n2−a2\sum_{n=1}^{\infty} \frac{1}{n^2-a^2} is: ∑n=1∞1n2−a2=12a2−πcot⁡(πa)2a\sum_{n=1}^{\infty} \frac{1}{n^2-a^2} = \frac{1}{2a^2} - \frac{\pi \cot(\pi a)}{2a} We can rewrite our sum SS by considering a related sum: ∑n=1∞136n2−1=∑n=1∞1(6n)2−12\sum_{n=1}^{\infty} \frac{1}{36n^2-1} = \sum_{n=1}^{\infty} \frac{1}{(6n)^2-1^2} Let a=1/6a = 1/6. Then a2=1/36a^2 = 1/36. ∑n=1∞1n2−(1/6)2=∑n=1∞3636n2−1=36S\sum_{n=1}^{\infty} \frac{1}{n^2 - (1/6)^2} = \sum_{n=1}^{\infty} \frac{36}{36n^2 - 1} = 36S Using the formula: 36S=12(1/6)2−πcot⁡(π/6)2(1/6)36S = \frac{1}{2(1/6)^2} - \frac{\pi \cot(\pi/6)}{2(1/6)} 36S=12(1/36)−πcot⁡(π/6)1/336S = \frac{1}{2(1/36)} - \frac{\pi \cot(\pi/6)}{1/3} 36S=18−3πcot⁡(π/6)36S = 18 - 3\pi \cot(\pi/6) Since cot⁡(π/6)=3\cot(\pi/6) = \sqrt{3}: 36S=18−3π336S = 18 - 3\pi \sqrt{3} Dividing by 36: S=18−3π336=12−π312S = \frac{18 - 3\pi \sqrt{3}}{36} = \frac{1}{2} - \frac{\pi \sqrt{3}}{12} We need to find S1+S2S_1 + S_2. S1=1⋅S=12−π312S_1 = 1 \cdot S = \frac{1}{2} - \frac{\pi \sqrt{3}}{12}. S2=2⋅S=2(12−π312)=1−π36S_2 = 2 \cdot S = 2 \left( \frac{1}{2} - \frac{\pi \sqrt{3}}{12} \right) = 1 - \frac{\pi \sqrt{3}}{6}. S1+S2=(12−π312)+(1−π36)S_1 + S_2 = \left( \frac{1}{2} - \frac{\pi \sqrt{3}}{12} \right) + \left( 1 - \frac{\pi \sqrt{3}}{6} \right) S1+S2=12+1−π312−2π312S_1 + S_2 = \frac{1}{2} + 1 - \frac{\pi \sqrt{3}}{12} - \frac{2\pi \sqrt{3}}{12} S1+S2=32−3π312S_1 + S_2 = \frac{3}{2} - \frac{3\pi \sqrt{3}}{12} S1+S2=32−π34S_1 + S_2 = \frac{3}{2} - \frac{\pi \sqrt{3}}{4} The given information ∑R=1∞1R2=π26\sum_{R=1}^{\infty} \frac{1}{R^2} = \frac{\pi^2}{6} is not directly used in this solution method but is a well-known result related to infinite series.