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The Kronecker Symbol $\left(\frac nm\right)$ has a range of $\{-1,0,1\}$ and $\sum\limits_{n=1}^\infty\frac{(-1)^n}n=-\ln(2)$, so we combine to find the following with the using software: $$\sum_{n=1}^\infty \frac {\left(\frac 2n\right)}n=\sum_{n=1}^\infty \frac {\left(\frac n2\right)}n=\frac{\si...