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

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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}

∑n=0∞(−1)n+1(2n−1)2=∑n=0∞−(−1)n4n2−4n+1=−∑n=0∞(−1)n4n2−4n+1   using the formula ∑n=k∞Cf(n)=C∑n=k∞f(n)=−∑n=0∞(−1)n(2n−1)2=−∑n=0∞(−1)n4n2−4n+1=−1+iLi⁡2(−i)2−iLi⁡2(i)2\begin{aligned}\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n + 1}}{\left(2 n - 1\right)^{2}} &= \sum_{n=0}^{\infty} -\frac{\left(-1\right)^{n}}{4 n^{2} - 4 n + 1}\\&= - \sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}}{4 n^{2} - 4 n + 1} \ \ \ \small{\color{grey}\text{using the formula } \sum_{n=k}^{\infty} C f{\left(n \right)} = C \sum_{n=k}^{\infty} f{\left(n \right)}}\\&= - \sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}}{\left(2 n - 1\right)^{2}}\\&= - \sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}}{4 n^{2} - 4 n + 1}\\&= -1 + \frac{i \operatorname{Li}_{2}\left(- i\right)}{2} - \frac{i \operatorname{Li}_{2}\left(i\right)}{2}\end{aligned}

ii is the imaginary unit, defined as i2=−1i^2 = -1.

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