\sum _{n=0}^{\infty }\:\left(-1\right)^{n+1}\frac{1}{\left(2n-1\right)^2}

asked by guest
on Nov 15, 2024 at 5:55 pm



You asked:

Evaluate the expression: ∑n=0∞(−1)n+1⋅1(2n−1)2\sum_{n = 0}^{\infty} {{\left( -1 \right)}^{n + 1} \cdot \frac{1}{{\left( 2 n - 1 \right)}^{2}}}

MathBot Answer:

∑n=0∞(−1)n+1(2n−1)2=−1+iLi⁡2(−i)2−iLi⁡2(i)2\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n + 1}}{\left(2 n - 1\right)^{2}} = -1 + \frac{i \operatorname{Li}_{2}\left(- i\right)}{2} - \frac{i \operatorname{Li}_{2}\left(i\right)}{2}

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