Interesting Eta????

First note that $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n(n+1)} = 2\ln2 - 1$ and that $\begin{aligned} S & = \; \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n(n+1)}\eta(n) \; = \; \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n(n+1)}\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m^n} \\ & = \; \sum_{m=1}^\infty (-1)^{m-1}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n(n+1)}\big(\tfrac{1}{m}\big)^n \; = \; 2\ln2 - 1 + \sum_{m=2}^\infty (-1)^{m-1}\sum_{n=1}^\infty (-1)^{n-1}\left[\tfrac{1}{n} - \tfrac{1}{n+1}\right]\big(\tfrac{1}{m}\big)^m \\ & = \; 2\ln2 - 1 + \sum_{m=2}^\infty(-1)^{m-1}\left\{\ln\big(1 + \tfrac{1}{m}\big) + \sum_{n=1}^\infty \frac{(-1)^n}{n+1}\big(\tfrac{1}{m}\big)^n\right\} \; = \; 2\ln2 - 1 + \sum_{m=2}^\infty (-1)^{m-1}\left\{\ln\big(1 + \tfrac{1}{m}\big) + m\left[\ln\big(1 + \tfrac{1}{m}\big) - \tfrac{1}{m}\right]\right\} \\ & = \; \sum_{m=1}^\infty (-1)^{m-1}\left\{(m+1)\ln\big(\tfrac{m+1}{m}\big) - 1\right\} \end{aligned}$ Note that $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{n(n+1)m^n} \; < \; \infty$ and so reversing the order of summation in the above argument is valid, and all series involved in the above calculation are absolutely convergent. Thus $S = \lim_{M \to \infty}S_M$ where $\begin{aligned} S_M & = \; \sum_{m=1}^{2M}(-1)^{m-1}\left\{(m+1)\ln\big(\tfrac{m+1}{m}\big) - 1 \right\} \; = \; \sum_{m=1}^{2M}(-1)^{m-1}(m+1)\ln(m+1) + \sum_{m=1}^{2M}(-1)^m(m+1)\ln m \\ & = \; \sum_{m=2}^{2M+1}(-1)^m m \ln m + \sum_{m=1}^{2M}(-1)^m(m+1)\ln m \; = \; -(2M+1)\ln(2M+1) + 2\sum_{m=1}^{2M}(-1)^m m \ln m + \sum_{m=1}^{2M} (-1)^m \ln m \end{aligned}$ Now $\begin{aligned} \sum_{m=1}^{2M} (-1)^m \ln m & = \; \ln\left(\frac{2 \times 4 \times \cdots \times 2M}{1 \times 3 \times \cdots \times (2M-1)}\right) \; = \; \ln\left(\frac{2^{2M}(M!)^2}{(2M!)}\right) \\ \sum_{m=1}^{2M} (-1)^m m \ln m & = \; \ln\left(\frac{2^2 \times 4^4 \times \cdots \times (2M)^{2M}}{1^1 \times 3^3 \times \cdots \times (2M-1)^{2M-1}}\right) \; = \; \ln\left(\frac{2^{2M(M+1)}P_M^4}{P_{2M}}\right) \end{aligned}$ where $P_M \; = \; \prod_{m=1}^M m^m$ Stirling's approximation tells us that $\frac{2^{2M}(M!)^2}{(2M)!} \; = \; \sqrt{\pi M}K_M$ where $\lim_{M \to \infty}K_M = 1$ . Moreover $P_M \; = \; M^{\frac12M^2 + \frac12M + \frac{1}{12}}e^{-\frac14M^2}Q_M$ where $\lim_{M \to \infty}Q_M = A$ , the Glaisher-Kinkelin constant. Substituting all this in gives $\lim_{M \to \infty}S_M \; = \; \lim_{M\to\infty}\ln\left(\frac{1}{2^{\frac{7}{6}}\big(1 + \frac{1}{2M}\big)^{2M+1}} \frac{Q_M^8}{Q_{2M}^2} \sqrt{\pi} K_M\right) \; = \; \ln\left(\frac{A^6\sqrt{\pi}}{2^{\frac76}e}\right)$ making the answer $1 + 1 + 2 + 7 + 6 + 1 + 6 = \boxed{24}$ .

$e^{\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \eta (n)}{n (n+1)}} \Rightarrow \frac{\sqrt{\pi } A^6}{2 \sqrt[6]{2} e}$

$1 + 1 + 2 + 7 + 6 + 1 + 6 \Rightarrow 24$