What value does $\dfrac{B}{x}$ approach?
Source: Select Problems of "Lumbreras Editors"
If I divide I get that:
$$\dfrac{B}{x}=\sqrt[3]{\dfrac{x}{x}.\dfrac{1}{x^2}\sqrt[5]{x^4\sqrt[9]{x^{24}\sqrt[17]{x^{240}...}}}}$$
$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{\dfrac{x^4}{x^4} \dfrac{1}{x^6}\sqrt[9]{x^{24}\sqrt[17]{x^{240}...}}}}$$
$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{ 1\sqrt[9]{\dfrac{x^{24}}{x^{24}}\dfrac{1}{x^{30}}\sqrt[17]{x^{240}...}}}}$$
$$\dfrac{B}{x}=\sqrt[3]{1\sqrt[5]{ 1\sqrt[9]{1\sqrt[17]{\dfrac{x^{240}}{x^{240}}\dfrac{1}{x^{270}}...}}}}$$ Can I assume that $\dfrac{B}{x}$ is approximately 1?
$B = x^\alpha$ where $$ \begin{align} \alpha &= {\small \frac13}(1 + {\small \frac15}(4 + {\small \frac19}(24 + {\small \frac{1}{17}}(240+\cdots ))))\\ &= \frac13 + \frac{4}{3\cdot 5} + \frac{24}{3\cdot 5 \cdot 9} + \frac{240}{3\cdot 5 \cdot 9\cdot 17} + \cdots\\ &= \frac{2}{2\cdot 3} + \frac{4}{3\cdot 5} + \frac{8}{5\cdot 9} + \frac{16}{9\cdot 17} + \cdots\\ &= \sum_{k=0}^\infty\frac{2^{k+1}}{(2^k+1)(2^{k+1}+1)}\\ &= \sum_{k=0}^\infty \left(\frac{2}{2^k+1} - \frac{2}{2^{k+1}+1}\right)\\ &= \frac{2}{2^0+1}\\ &= 1 \end{align} $$ This means $\frac{B}{x} = \frac{x}{x} = 1$.