Intriguing Sum

We have $\sum_{n=0}^\infty \sum_{k=0}^n 2^{-n} (-1)^k {n \choose k}x^k \; = \; \sum_{n=0}^\infty 2^{-n}(1 - x)^n \; = \; \frac{2}{1+x} \;,$ and this series is absolutely convergent for $-1 < x \le 1$ and uniformly convergent for $0 \le x \le 1$ . Since $\int_0^1 x^\alpha (\ln x)^3\,dx \; = \; \frac{-6}{(\alpha+1)^4} \;, \qquad \alpha > -1 \;,$ we see that $S \; = \; -\tfrac16\int_0^1 \frac{2}{1+x} (\ln x)^3\,dx \; = \; -\tfrac13\sum_{n=0}^\infty (-1)^n \int_0^1 x^n(\ln x)^3\,dx \; = \; 2\sum_{n=0}^\infty (-1)^n \frac{1}{(n+1)^4}$ and so $S \; = \; 2\left(\sum_{n=1}^\infty \frac{1}{n^4} - 2\sum_{n=1}^\infty \frac{1} {(2n)^4}\right) \; = \; 2\big(1 - \tfrac18)\zeta(4) \; = \; \tfrac{7}{360}\pi^4 \;,$ making the answer $7 + 360 + 4 = \boxed{371}$ .

Using Euler's (backward) Series Transformation , with $a_k=\frac{1}{(k+1)^4}$ , we find that $S=2\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^4}$ $=2\eta(4)=2\left(1-\frac{1}{8}\right)\zeta(4)=\frac{7}{360}\pi^4$ . The relationship between the Dirichlet Eta Function $\eta(s)$ and the Zeta Function $\zeta(s)$ was discussed here .