Evaluar

Question

¿Cómo podemos evaluar:

ps

Intenté transformarlo en$$\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$$\sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$ = \ ln \ left (\ sqrt {\ frac {1} {\ left (\ sqrt {2 ^ {n 1} 2} \ sqrt {2 ^ {n 2} 2} \ right) ^ 2} 1} \ frac {1} {\ sqrt {2 ^ {n 1} 2} \ sqrt {2 ^ {n 2 } 2}} \ right) $$ Para que la expresión original se convierta en:$$$ $$$\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$ $ Pero no se ve bien en absoluto.
La respuesta final está cerca de 0.658479, que es$$=\ln\left(\prod_{n=1}^\infty\left(\sqrt{\frac{1}{\left(\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}\right)^2}+1}+\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)\right)$

Answer 1

Tenga en cuenta que $$ \ sinh ^ {- 1} x- \ sinh ^ {- 1} y = \ sinh ^ {- 1} \ left (x \ sqrt {1 y ^ 2} -y \ sqrt {1 ^ 2} \ right) $$

\begin{align*} \frac{1}{\sqrt{2^{n+2}+2}+\sqrt{2^{n+1}+2}} &= \frac{\sqrt{2^{n+2}+2}-\sqrt{2^{n+1}+2}} {(2^{n+2}+2)-(2^{n+1}+2)} \\ &= \frac{\sqrt{2^{n+2}+2}-\sqrt{2^{n+1}+2}} {2^{n+1}} \\ &= \sqrt{\frac{1}{2^{n}} \left( 1+\frac{1}{2^{n+1}} \right)}- \sqrt{\frac{1}{2^{n+1}} \left( 1+\frac{1}{2^{n}} \right)} \\ \sinh^{-1} \left( \frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}} \right) &= \sinh^{-1} \frac{1}{\sqrt{2^{n}}}-\sinh^{-1} \frac{1}{\sqrt{2^{n+1}}} \\ \sum_{n=1}^{\infty} \sinh^{-1} \left( \frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}} \right) &= \sinh^{-1} \frac{1}{\sqrt{2}} \\ &= \ln \left( \frac{1+\sqrt{3}}{\sqrt{2}} \right) \\ &= \ln \sqrt{2+\sqrt{3}} \end{align*}