\sum _{n=0}^{\infty }\:\left(-1\right)^{n+1}\frac{1}{\left(2n-1\right)^2}
You asked:
Evaluate the expression: \(\sum_{n = 0}^{\infty} {{\left( -1 \right)}^{n + 1} \cdot \frac{1}{{\left( 2 n - 1 \right)}^{2}}}\)
MathBot Answer:
\[\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n + 1}}{\left(2 n - 1\right)^{2}} = -1 + \frac{i \operatorname{Li}_{2}\left(- i\right)}{2} - \frac{i \operatorname{Li}_{2}\left(i\right)}{2}\]