Prove-by-induction-on-n-for-n-2-u-n-2-3-n-1-for-the-sequence-u-n-defined-by-the-recurrence-relation-u-1-1-

Question Number 591 by 112358 last updated on 04/Feb/15

P r o v e b y i n d u c t i o n o n n, f o r n ⩾ 2,

u_{n} ⩾ 2^{3^{n - 1}}

f o r t h e s e q u e n c e {u_{n}} d e f i n e d b y

t h e r e c u r r e n c e r e l a t i o n

u_{1} = 1

u_{n + 1} = {(u_{n} + \frac{1}{u_{n}})}^{3}, n ⩾ 1 .

Answered by prakash jain last updated on 04/Feb/15

u_{n + 1} = {(u_{n} + \frac{1}{u_{n}})}^{3} ⩾ u_{n}^{3}

If u_{n} ⩾ 2^{3^{n - 1}}

u_{n + 1} ⩾ {(2^{3^{n - 1}})}^{3} = 2^{3^{n - 1}} \times 2^{3^{n - 1}} \times 2^{3^{n - 1}} = 2^{3^{n}}

or

u_{n} ⩾ 2^{3^{n - 1}} \Rightarrow u_{n + 1} ⩾ 2^{3^{n}}

u_{2} = 2^{3} ⩾ 2^{3^{2 - 1}} \Rightarrow u_{3} ⩾ 2^{3^{2}}