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$\begin{array}{ll} &\left(1-2^{1-s}\right)\zeta(s)\\[2mm] =&\left(1-2^{1-s}\right)\sum_{n=1}^\infty\frac{1}{n^s}\\[2mm] =&\sum_{n=1}^\infty\frac{1}{n^s}-2^{1-s}\sum_{n=1}^\infty\frac{1}{n^s}\\[2mm] =&\sum_{n=1}^\infty\frac{1}{n^s}-2\sum_{n=1}^\infty\frac{1}{(2n)^s}\\[2mm] =&\left(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\ldots\right)-2\left(\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+\ldots\right)\\[2mm] =&\frac{1}{1^s}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\ldots\\[2mm] =&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\\[2mm] =&\sum_{n=0}^\infty\frac{(-1)^{n}}{(n+1)^s} \end{array}$

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$\begin{array}{rcl} \zeta(s)&=&\frac{1}{1-2^{1-s}}\sum_{k=0}^\infty\frac{(-1)^{k}}{(k+1)^s}\\[2mm] &=&\frac{1}{1-2^{1-s}}\sum_{n=0}^\infty\frac{1}{(1+y)^{n+1}}\sum_{k=0}^n{n\choose k}y^{k+1}\frac{(-1)^{k}}{(k+1)^s}\\[2mm] &=&\frac{1}{1-2^{1-s}}\sum_{n=0}^\infty\frac{1}{(1+1)^{n+1}}\sum_{k=0}^n{n\choose k}1^{k+1}\frac{(-1)^{k}}{(k+1)^s}\\[2mm] &=&\frac{1}{1-2^{1-s}}\sum_{n=0}^\infty\frac{1}{2^{n+1}}\sum_{k=0}^n{n\choose k}\frac{(-1)^{k}}{(k+1)^s}\\[2mm] \end{array}$

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$\begin{array}{rcl} \sum_{k=0}^\infty A_k&=&\sum_{k=0}^\infty \left(\sum_{n=k}^\infty {n\choose k}\frac{y^{k+1}}{(1+y)^{n+1}}\right)A_k\\[2mm] &=&\sum_{k=0}^\infty \sum_{n=k}^\infty {n\choose k}\frac{y^{k+1}}{(1+y)^{n+1}}A_k\\[2mm] &=&\sum_{n=0}^\infty \sum_{k=0}^n {n\choose k}\frac{y^{k+1}}{(1+y)^{n+1}}A_k\\[2mm] &=&\sum_{n=0}^\infty \frac{1}{(1+y)^{n+1}}\sum_{k=0}^n {n\choose k}y^{k+1}A_k\\[2mm] \end{array}$

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$\begin{array}{rcl} \zeta(-1)&=&\frac{1}{1-2^{2}}\sum_{n=0}^\infty\frac{1}{2^{n+1}}\sum_{k=0}^n{n\choose k}(-1)^{k}(k+1)\\[2mm] &=&-\frac{1}{3}\sum_{n=0}^1\frac{1}{2^{n+1}}\sum_{k=0}^n{n\choose k}(-1)^{k}(k+1)\\[2mm] &=&-\frac{1}{3}\cdot\frac{1}{2^{0+1}}{0\choose 0}(-1)^{0}(0+1)-\frac{1}{3}\cdot\frac{1}{2^{1+1}}{1\choose 0}(-1)^{0}(0+1)-\frac{1}{3}\cdot\frac{1}{2^{1+1}}{1\choose 1}(-1)^{1}(1+1)\\[2mm] &=&-\frac{1}{3}\cdot\frac{1}{2}\cdot1\cdot1\cdot1-\frac{1}{3}\cdot\frac{1}{4}\cdot1\cdot1\cdot1-\frac{1}{3}\cdot\frac{1}{4}\cdot1\cdot(-1)\cdot2\\[2mm] &=&-\frac{1}{6}-\frac{1}{12}+\frac{1}{6}\\[2mm] &=&-\frac{1}{12} \end{array}$

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$\begin{array}{rl} &\sum_{k=n}^\infty{k\choose n}\frac{1}{(1+y)^{k+1}}\\[2mm] =&{n\choose n}\sum_{k=n}^\infty\frac{1}{(1+y)^{k+1}}+\sum_{k=n}^\infty\left({k+1\choose n}-{k\choose n}\right)\sum_{j=k+1}^\infty \frac{1}{(1+y)^{j+1}}\\[2mm] =&\sum_{k=n}^\infty\frac{1}{(1+y)^{k+1}}+\sum_{k=n}^\infty{k\choose n-1}\sum_{j=k+1}^\infty \frac{1}{(1+y)^{j+1}}\\[2mm] =&\frac{\frac{1}{(1+y)^{n+1}}}{1-\frac{1}{(1+y)}}+\sum_{k=n}^\infty{k\choose n-1}\frac{\frac{1}{(1+y)^{k+2}}}{1-\frac{1}{(1+y)}}\\[2mm] =&\frac{1}{y(1+y)^n}+\sum_{k=n}^\infty{k\choose n-1}\frac{1}{y(1+y)^{k+1}}\\[2mm] =&\sum_{k=n-1}^\infty{k\choose n-1}\frac{1}{y(1+y)^{k+1}}\\[2mm] =&\cdots\\[2mm] =&\sum_{k=0}^\infty{k\choose 0}\frac{1}{y^n(1+y)^{k+1}}\\[2mm] =&\frac{1}{y^n}\sum_{k=0}^\infty\frac{1}{(1+y)^{k+1}}\\[2mm] =&\frac{1}{y^n}\frac{\frac{1}{(1+y)}}{1-\frac{1}{(1+y)}}\\[2mm] =&\frac{1}{y^{n+1}} \end{array}$

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