Benedict Irwin edited untitled.tex  over 8 years ago

Commit id: 56feff48e44b2b98391496dada22da5a40930172

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It seems clear that a factor of $8^{n+1}$ must be removed from the denominator in the sequence.  \begin{align}  1-\frac{41\cdot4}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \frac{1}{2}+\frac{2816}{3^3\cdot8^3\sqrt{3}\pi}-\frac{41\cdot2^2}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \frac{1}{3}-\frac{272753}{3\cdot4^4\cdot8^4\sqrt{4}\pi^{3/2}}+2\frac{2816}{3^3\cdot8^3\sqrt{3}\pi}-\frac{41\cdot2^2}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \frac{1}{4}+\frac{65209\cdot8^5}{500\cdot8^5\cdot5^5\sqrt{5}\pi^2}-3\frac{272753}{3\cdot4^4\cdot8^4\sqrt{4}\pi^{3/2}}+3\frac{2816}{3^3\cdot8^3\sqrt{3}\pi}-\frac{41\cdot2^2}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \frac{1}{5}-\frac{61710899}{2866544640\sqrt{6}\pi^{5/2}}+4\frac{65209}{1562500\sqrt{5}\pi^2}-6\frac{272753}{3145728\sqrt{4}\pi^{3/2}}+4\frac{11}{54\sqrt{3}\pi}-\frac{41}{64\sqrt{2}\pi^{1/2}}\\ 1-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \frac{1}{2}+\frac{2816}{3^3\cdot8^3\sqrt{3}\pi}-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \frac{1}{3}-\frac{272753}{3\cdot4^4\cdot8^4\sqrt{4}\pi^{3/2}}+2\frac{2816}{3^3\cdot8^3\sqrt{3}\pi}-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \frac{1}{4}+\frac{65209\cdot8^5}{500\cdot8^5\cdot5^5\sqrt{5}\pi^2}-3\frac{272753}{3\cdot4^4\cdot8^4\sqrt{4}\pi^{3/2}}+3\frac{2816}{3^3\cdot8^3\sqrt{3}\pi}-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \frac{1}{5}-\frac{61710899}{2866544640\sqrt{6}\pi^{5/2}}+4\frac{65209}{1562500\sqrt{5}\pi^2}-6\frac{272753}{3145728\sqrt{4}\pi^{3/2}}+4\frac{11}{54\sqrt{3}\pi}-\frac{164}{2^2\cdot8^2\sqrt{2}\pi^{1/2}}\\  \vdots\\  s_2=\sum_{n=1}^\infty\frac{1}{n} + \sum_{k=0}^{n-1} \begin{pmatrix}n-1\\k\end{pmatrix}\frac{(-1)^{n-k}a_k}{b_k(k+2)^{(k+2)}\sqrt{k+2}\pi^{(k+1)/2}} \begin{pmatrix}n-1\\k\end{pmatrix}\frac{(-1)^{n-k}a_k}{8^{(k+2)}b_k(k+2)^{(2k+5)}\pi^{(k+1)/2}}  \end{align}