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Answer 1 · 2015-09-25T17:32:16

It is divergent

All the terms of this sequence are greater than one. Hence as #n-> oo, a_n !=0#. Sequence does not therefore converge. It diverges.

Answer 2 · 2015-09-25T17:43:45

The sequence converges to #2#.

The sequence is increasing and bounded (by 2) so it must converge.

Let #b_n = a_(n+1)#, i.e. #b_n=sqrt(2+a_n)#

#b_n# must converge to the same limit as #a_n#, so, calling the limit #L#, we have #L = sqrt(2+L)#.

Solving the equation gets us #L=2#.