Verified answer

a( 1 ) = 2

a ( 2 ) = 4/3 which is less than 2

So assume that a ( n ) < a ( n - 1 ) < a ( n - 2 < ... < a ( 1 )

Now use that to show that that a ( n + 1 ) < a ( n )

we know that a ( n ) < a ( n - 1 )

then if we square both sides and divide by 1/3 we get

1/2 ( a ( n ) )^2 < 1/3 ( a ( n -1 ))^2

which means a ( n + 1 ) < a ( n )

so via mathematical induction we've shown that your sequence is constantly decreasing, with an upper bound of 2.

and since a ( n + 1 ) = 1/3(an)^2 is always a positive number, that means a lower bound is 0.

If you have a sequence which constantly decreasing with a lower bound, then you know that a limit exists.

If a limit exists, then as n -> infinite, a (n + 1 ) and a ( n ) will approach the same value. call it x

so, let n -> infinite you get

x = 1/3 x^2

0 = x ( 1/3x - 1 )

so, x = 0, or x = 3. Since we know our upper bound is 2, then we disregard the answer 3.

So 0 is the limit of your sequence.