$\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n+1 } } \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) } \frac { { \left( -1 \right) }^{ k } }{ k+1 } = \frac {\pi^a}b$

The equation above holds true for integers $a$ and $b.$ Submit your answer as $a + b$ .

The answer is 8.

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Since $\sum_{k=0}^{n}{n\choose k}\frac{(-1)^k}{k+1} =\int_0^1\left(\sum_{k=0}^n{n\choose k} (-x)^k\right)dx=\int_0^1(1-x)^ndx=\int_0^1 x^ndx=\frac{1}{n+1}$ which give the final closed form $\sum_{n\geq 0}\frac{1}{(n+1)^2}=\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}$

The finite sum can be observed as follows by splitting the sum in even and odd parity $f_{k\geq 0}(n)=\sum_{k\leq n}{n\choose k}\frac{(-1)^k}{k+1} =\sum_{k\leq n}\left({n\choose 2k}\frac{1}{2k+1}-{n\choose 2k+1}\frac{1}{2k+2}\right)=\sum_{k\leq n}\left({n+1\choose 2k+1}-{n+1\choose 2k+2}\right)\frac{1}{n+1}=\frac{1}{n+1}$ Alternative way to go is explained by Carsten Meyer in comment section.