liouville tau

Since $\sum_{k=0}^\infty (-1)^k (k+1)x^k \; = \; (1 + x)^{-2} \qquad |x| < 1$ we have $\sum_{n=1}^\infty \frac{\lambda(n)\tau(n)}{n^2} \; = \; \prod_{p} \left(1 + \tfrac{1}{p^2}\right)^{-2} \; = \; \left(\frac{15}{\pi^2}\right)^{-2} \; = \; \frac{\pi^4}{225}$

Consider a completely multiplicative function, f(n), then $(f*f)(n)=\sum_{d|n} f(d)f(n/d)=\sum_{d|n} f(n)=f(n)\tau(n)$ in this case, $\lambda(n)$ is a multiplicative function. We put this in dirichlet series to get $\left(\sum_{n=1} \dfrac{\lambda(n)}{n^2}\right)^2=\sum_{n=1}^\infty \dfrac{\lambda(n)\tau(n)}{n^2}$ We know the dirichlet series for $\sum_{n=1} \dfrac{\lambda(n)}{n^s}=\dfrac{\zeta(2s)}{\zeta(s)}$ So we have $\sum_{n=1}^\infty \dfrac{\lambda(n)\tau(n)}{n^2}=\dfrac{\zeta^2(4)}{\zeta^2(2)}=\dfrac{\pi^4}{225}$