Very fancy $\zeta(n)$ sum

A slightly different proof. For any prime $p$ and real $s > 0$ define $G_{p,s}(t) \; = \; \sum_{k=0}^\infty \frac{e^{-tH_k}}{p^{ks}} \hspace{2cm} t \ge 0$ (recall that $H_0=0$ ). Then $G_{p,s}$ is differentiable, and $\begin{aligned} G_{p,s}(0) & = \; \sum_{k=0}^\infty p^{-ks} \; = \; \big(1 - p^{-s}\big)^{-1} \\ G_{p,s}'(0) & = \; -\sum_{k=0}^\infty \frac{H_k}{p^{ks}} \; = \; -\sum_{k=1}^\infty \sum_{j=1}^k \frac{1}{jp^{ks}} \; = \; -\sum_{j=1}^\infty \frac{1}{jp^{js}} \sum_{k=j}^\infty \frac{1}{p^{(k-j)s}} \\ & = \; -\sum_{j=1}^\infty \frac{1}{jp^{(j-1)s}}\big(1 - p^{-s}\big)^{-1} \; = \; \big(1 - p^{-s}\big)^{-1}\ln\big(1 - p^{-s}\big) \end{aligned}$ and hence $\frac{G_{p,s}'(0)}{G_{p,s}(0)} \; = \; \ln\big(1 - p^{-s}\big)$ Now $\chi$ is additive, and hence $X_t(n) = e^{-t\chi(n)}$ is multiplicative. Thus, if we define $F_s(t) \; = \; \sum_{n=1}^\infty \frac{X_t(n)}{n^s} \hspace{2cm} t \ge 0$ then $F_s(t) \; = \; \prod_p\left(\sum_{k=0}^\infty \frac{X_t(p^k)}{p^{ks}}\right) \; = \; \prod_p G_{p,s}(t)$ so that $\begin{aligned} F_s(0) & = \; \prod_pG_{p,s}(0) \; = \; \prod_p\big(1 - p^{-s}\big)^{-1} \; = \; \zeta(s) \\ \frac{F_s'(t)}{F_s(t)} & = \; \sum_p \frac{G_{p,s}'(t)}{G_{p,s}(t)} \\ \frac{F_s'(0)}{F_s(0)} & = \; \sum_p \frac{G_{p,s}'(0)}{G_{p,s}(0)} \; = \; \sum_p \ln\big(1 - p^{-s}\big) \; = \; \ln\left(\prod_p\big(1 - p^{-s}\big)\right) \; = \; -\ln \zeta(s) \end{aligned}$ and hence $F_s'(0) = -\zeta(s)\ln\zeta(s)$ .But this implies that $\sum_{n=1}^\infty \frac{\chi(n)}{n^s} \; = \; \zeta(s)\ln\zeta(s) \hspace{2cm} s > 0$ In the case $n=4$ we obtain the sum $\tfrac{1}{90}\pi^4 \ln\big(\tfrac{1}{90}\pi^4\big) \; = \; \pi^4 \ln\left(\sqrt[90]{\frac{\pi^4}{90}}\right)$ making the answer $4+90+4+90=\boxed{188}$ .

$\large \displaystyle \chi(n) = \sum_{j =1}^i H_{k_j} = \sum_{d|n}\frac{\Lambda(d)}{\log d} = \frac{\Lambda(n)}{\log n} * 1 \\\\ \large \displaystyle \sum_{n=2}^{\infty} \frac{\chi(n)}{n^4} = \sum_{n=1}^{\infty}\frac{1}{n^4} \sum_{n=2}^{\infty} \frac{\Lambda(n)}{\log n} \frac{1}{n^4} = \zeta(4)\log \zeta(4) = \frac{\pi^4}{90} \log \left( \frac{\pi^4}{90} \right)$

Note: See first comment and it's reply for some visual help and more details.

From the definition, we can see $\chi(n)$ is additive . This means we can group the multiples of each prime and break the sum into:

$\displaystyle \sum_{p \text{ prime}} \sum_{i \geq 1} \frac{\chi(p^k)}{(pi)^4}$

Where $k$ is the exponent of $p$ in the prime factorization of $pi$ . Also, $\chi(p^k)=H_k$ , which allows us to break this sum even further into:

$\displaystyle \sum_{p \text{ prime}} \sum_{k \geq 1} \sum_{j \geq 1} \frac{1}{k(p^{k}j)^4}$

$\displaystyle \sum_{p \text{ prime}} \sum_{k \geq 1} \frac{1}{kp^{4k}} \sum_{j \geq 1} \frac{1}{j^4} = \zeta(4) \sum_{p \text{ prime}} \sum_{k \geq 1} \frac{1}{kp^{4k}}$

From the taylor expansion of $\ln(1-x)$ we get:

$- \zeta(4) \! \! \! \sum_{p \text{ prime}} \! \! \ln \! \left( \! 1 - \frac{1}{p^4} \! \right) \! \! = \! \! - \zeta(4) \ln \! \left( \! \prod_{p \text{ prime}} \! \! \left( 1 - \frac{1}{p^4} \! \right) \! \! \right)$ $- \zeta(4) \ln \! \left( \frac{1}{\zeta(4)} \! \right) = \zeta(4) \ln \! \left( \zeta(4) \right)$

Which gives us:

$\displaystyle \frac{\pi^4}{90}\ln\left( \frac{\pi^4}{90}\right) = \pi^4\ln \! \left( \! \sqrt[ \small 90]{\frac{\pi^4}{90}}\right)$

And the sum is $90+90+4+4 = \boxed{188}$

Bonus: In general $\displaystyle \sum_{n \geq 2} \frac{{\chi(n)}}{n^s} = \zeta(s)\ln(\zeta(s))$