Harder Nested Radicals

Thanks for noticing the typo; all corrected now. :)

$\sqrt { 2\sqrt [ 3 ]{ 2\sqrt [ 4 ]{ 2\sqrt [ 5 ]{ 2... } } } } =\sqrt { 2 } \times \sqrt [ 2\times 3 ]{ 2 } \times \sqrt [ 2\times 3\times 4 ]{ 2 } ...$ $={ 2 }^{ \frac { 1 }{ 2 } }{ \times 2 }^{ \frac { 1 }{ 2\times 3 } }\times { 2 }^{ \frac { 1 }{ 2\times 3\times 4 } }...$ $={ 2 }^{ (\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ k! } }) }={ 2 }^{ e-2 }$

$^{ \sum _{ k=0 }^{ \infty }{ \frac { 1 }{ k! } } }=e$ (definition of $e$ )

Hope this helped.

It would be easier if you also said which part you didn't understand.

Yes. Euler's number , $e$ , is equivalent to $\displaystyle\sum_{k=0}^{\infty} \dfrac{1}{k!}$ .

By definition $e = \lim_{x \rightarrow \infty} (1 + \frac{1}{x})^{x}$ .

The fact that this the above series and limit have exactly the same value is pretty amazing, really. :)