Proving a decreasing, convergent sequence.

Question

Let $t_1=4$ and $t_{n+1}= \sqrt{3+2t_n}$. How do I prove that $3<t_{n+1}<t_n$? I understand that it is true, but I don't understand how to show it...

Thanks!

Answer 1

It is obvious that t_1 = 4 > 3.

Assuming $ t_k > 3 $, then

$ t_{k+1}^2 = 3 + 2t_k > 3 + 2 \times 3 > 9 $

That implies $ t_{k+1} > 3 $

By principle of mathematical induction, $ t_n > 3 $ for all $ n \ge 1 $

After establishing the fact above, we can prove $ t_n > t_{n-1} $ easily as follow:

$ t_n > t_{n+1} $

$ t_n > \sqrt{3 + 2t_n} $

$ t_n^2 > 3 + 2t_n $

$ t_n^2 - 2t_n > 3 $

$ t_n^2 - 2t_n + 1 > 4 $

$ (t_n- 1)^2 > 4 $

$ (t_n- 1) > 2 $

$ t_n > 3 $

Of course, you read this backwards!

Answer 2

It suffices to note that, if $t>3$, then $3<\sqrt{3+2t}<t$.

The first inequality proves all terms in the sequence are $>3$, the second inequality proves the sequence is decreasing.