Ramanujan stated this radical in his lost notebook:

√5+√5+√5−√5+√5+√5+√5−⋯=2+√5+√15−6√52

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

Any help appreciated.

Thanks.

**Answer**

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

x1=+√5+√5+√5−√5+√5+√5+√5−⋯=2+√5+√15−6√52=2.7472…

The other roots of the quartic in x are given by the patterns (+,+,−,+),(+,−,+,+),(−,+,+,+), respectively

x2=+√5+√5−√5+√5+√5+√5−√5+⋯=2−√5+√15+6√52=2.5473…

x3=+√5−√5+√5+√5+√5−√5+√5+⋯=2+√5−√15−6√52=1.4888…

x4=−√5+√5+√5+√5−√5+√5+√5+⋯=2−√5−√15+6√52=−2.7833…

This immediately implies that the four roots obey the system,

x21=x2+5x22=x3+5x23=x4+5x24=x1+5

also studied by Ramanujan. (See this related post.) More generally, using any of the 24=16 possible periods,

x=±√a±√a±√a±√a±…

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

x=(((x2−a)2−a)2−a)2−a

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

(x2−x−a)(x2+x−a+1)=0

and the other three had coefficients in the cubic,

y3+3y=4(1+ay)

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,

xn=−y−z4±12√(y−2)(y+z)z2y

where,

z=±√y2+4

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 Q, hence no cubic irrationalities are involved, and one of the xn 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 2n 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.

**Attribution***Source : Link , Question Author : Shobhit , Answer Author : Community*