The Power! series

Note: $\chi (n)$ is actually the Liouville function , typically represented as $\lambda (n)$ .

Since $\lambda(n)$ is completely multiplicative , we can factor $S$ :

$S=\left(\frac{1}{\left(2^0\right)^4}-\frac{1}{\left(2^1\right)^4}+\frac{1}{\left(2^2\right)^4}+...\right)\left(\frac{1}{\left(3^0\right)^4}-\frac{1}{\left(3^1\right)^4}+\frac{1}{\left(3^2\right)^4}+...\right)\left(\frac{1}{\left(5^0\right)^4}-\frac{1}{\left(5^1\right)^4}+\frac{1}{\left(5^2\right)^4}+...\right)...$

Using the sum of a geometric progression and the Euler product formula :

$\displaystyle S=\prod_{\text{primes}}\left(\frac{1}{1+\frac{1}{p^4}}\right)=\prod_{\text{primes}}\frac{1}{\left(1+\frac{1}{p^4}\right)}\cdot\frac{\left(1-\frac{1}{p^4}\right)}{\left(1-\frac{1}{p^4}\right)}=\frac{ \displaystyle \prod_{\text{primes}}\left(1-\frac{1}{p^4}\right)}{ \displaystyle \prod_{\text{primes}}\left(1-\frac{1}{p^8}\right)}=\frac{\zeta (8)}{\zeta(4)}=\frac{\pi^8}{9450}\cdot\frac{90}{\pi^4}=\frac{\pi^4}{105}$

Thus, the sum is $105 + 4 = \boxed{109}$

Slightly different approach:

Firstly, I define $\lambda(n)=(-1)^{\Omega(n)}$ to be Liouville function (as defined here ). Your definition is a bit faulty as Liouville function can be either $1$ or $-1$ , then $(-1)^{\lambda(n)}$ would be always $-1$ .

If $S$ is multiplied by function $\zeta(4)=\sum_{n=1}^{\infty}\frac{1}{n^4}$ , we get

$G= S \times \zeta(4) = \sum_{n=1}^{\infty} \frac{\lambda * 1(n)}{n^4}=\sum_{n=1}^{\infty}\frac{\sum_{d|n}\lambda (d) }{n^4}$

Note that $*$ indicates Dirichlet convolution. A property of Liouville function is

$\lambda(n)=\begin{cases} 1 & n \ perfect \ square\\ 0 & otherwise \\ \end{cases}$

Therefore $G=\sum_{n=1}^{\infty}\frac{1}{(n^2)^4}=\zeta(2\times 4)=\zeta(8)$

finally

$G=\zeta(8)=S\times \zeta(4) \implies S=\frac{\zeta(8)}{\zeta(4)}=\frac{\frac{\pi^8}{9450}}{\frac{\pi^4}{90}}$

Let, $\Lambda$ is a function of $s$ . Now,

$\Lambda(s)=1-\frac{1}{2^{s}}-\frac{1}{3^{s}}+\frac{1}{4^{s}} -\frac{1}{5^s}+\frac{1}{6^s} -\frac{1}{7^s}-\frac{1}{8^s}+\frac{1}{9^s}+.........+\frac{1}{15^s}+\frac{1}{16^s}-\frac{1}{17^s} -\frac{1}{18^s}-........$ .

Notice, $\Lambda=\sum \frac{(-1)^{\lambda(n)}}{n^s}$ . Where, $\lambda(n)=\sum_{n} a_i$ . Where $a_i$ s are the power of prime elements of $n$ .

Or, $\lambda(n)$ is the sum of the powers of the prime factors of n

For example , $\lambda(500)={2^2}{5^3}=2+3=5$ .

See, $(1+\frac{1}{2^s})\Lambda(s)=1-\frac{1}{3^s} -\frac{1}{5^s} -......$ .

And continuing like this we get....

$\Lambda(s)=\frac{1}{\prod_{primes}(1+\frac{1}{p^s})}$ .

Likely we get $\Lambda(s)=\frac{\zeta(2s)}{\zeta(s)}$ .

Replacing $s$ by $4$ in above equation we get $\Lambda(4)=\frac{\zeta(8)}{\zeta(4)}=\frac{π^4}{105}$ .

Hence, $a+c=109$ .