# How to prove √5+√5+√5−√5+√5+√5+√5−⋯=2+√5+√15−6√52\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}

Ramanujan stated this radical in his lost notebook:

I don’t have any idea on how to prove this.

Any help appreciated.

Thanks.

The correct period has length 4, namely (+,+,+,-)

The other roots of the quartic in $x$ are given by the patterns $\small(+,+,-,+),\; (+,-,+,+),\; (-,+,+,+)$, respectively

This immediately implies that the four roots obey the system,

also studied by Ramanujan. (See this related post.) More generally, using any of the $2^4=16$ possible periods,

will be the absolute value of a root of the 16th deg eqn,

In his Notebooks IV (p.42-43), Ramanujan stated that (1) was a product of 4 quartic polynomials, one of which is the reducible,

and the other three had coefficients in the cubic,

Using Mathematica to factor (1), we find that it is indeed a product of (2) and a 12th deg eqn with coefficients in a. After some manipulation, the 12 roots are,

where,

Since there are 4 sign changes and (3) gives 3 choices for $y$, this yields the 12 roots.

Note: For $a=5$ (as well as $a=2$), the cubic factors over $\mathbb{Q}$, hence no cubic irrationalities are involved, and one of the $x_n$ will give the value of the appropriate infinite nested radical.

P.S. Interestingly, for period length $n> 4$, not all the roots of the deg $2^n$ equation will be expressible as finite radical expressions for general $a$. The exception is $a=2$ where the solution involves roots of unity as discussed in this post.