I want to convert $\frac{\pi^2}{6}$ to continued fraction. I want to use the theorem that $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{\alpha_{k}} = \frac{1}{\alpha_{1}}_{+} \frac{\alpha_{1}^{2}}{\alpha_{2} - \alpha_{1}}_{+} \frac{\alpha_{2}^{2}}{\alpha_{3} - \alpha_{2}}_{+}\frac{\alpha_{3}^{2}}{\alpha_{4} - \alpha_{3}}_{+}\ldots $$But this needs an alternating series. However, I'm confused about how to express $\frac{\pi^{2}}{6}$ as an alternating series.
And my eventual goal is to get the continued fraction for $\frac{6}{\pi^2}$, which I think i could get from the continued fraction for $\frac{\pi^2}{6}$
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Miranda
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2You might be motivated by this often duplicated fact, the sum of inverse squares equalis $\pi^2/6$. We can shoehorn that into your scheme by replacing $\alpha_k$ by $-\alpha_k$ for odd indexes $k$. Of course the scheme was never designed to give simple continued fractions (as described here). – hardmath Oct 15 '23 at 00:37
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@hardmath This is really helpful. – Miranda Oct 15 '23 at 00:43
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1The power of $-1$ in your alternating series seems to be off-by-one. Compare this previous Answer which has $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{a_{n}}$$/ – hardmath Oct 15 '23 at 13:55
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@hardmath Yeah you're right – Miranda Oct 16 '23 at 02:54