Calling

#y = log_2(2+1/log_2(2+1/log_2(2+...)))# we have

#y = log_2(2+1/y)#

Now, calling

#y_{k+1} = log_2(2+1/y_k)#

substituting

#y_1 = 1.428# we obtain

#y_2=1.433# and sucessively

#y_3 =1.432# etc.

converging to

#(
(y_1 = 1.428000000000000000000000000000),
(y_2 = 1.433109072200861922041781326700),
(y_3 = 1.431774625761248698196936857130),
(y_4 = 1.432122371909281044466411003668),
(y_5 = 1.432031697667491831893471498920),
( cdots),
(y_21 = 1.432050448448111801269533316372),
(y_22 = 1.432050448448122681455174642906),
(y_23 = 1.432050448448119794875310617499),
(y_24 = 1.432050448448120683053730317624),
(y_25 = 1.432050448448120238964520467562),
(y_26 = 1.432050448448120238964520467562)
)#