# I think of it as hard proof by induction

1. May 30, 2008

### rock.freak667

[SOLVED] I think of it as hard proof by induction

1. The problem statement, all variables and given/known data

given for the sequence $a_1,a_2,a_3,a_4,...$,$a_1=1$ and that

$$a_{n+1}=(a_n+\frac{1}{a_n})^{\lambda}$$

where$\lambda>1$.Prove by mathematical induction that,for $n \geq 2$

$$a_n \geq 2^{g(n)}$$

where $g(n)=\lambda^{n-1}$

Prove also,for $n \geq 2$

$$\frac{a_{n+1}}{a_n}>2^{(\lambda -1)g(n)}$$
2. Relevant equations

3. The attempt at a solution

well what I tried to do was to consider $a_{n+1}-2^{g(n)}$

which after simplification gives:

$$\frac{(a_n^2+1)^{\lambda}}{a_n^{\lambda}}-2^{g(n)}[/itex] which I then said should be greater than or equal to zero since the fraction on the left is +ve and 2^g(n) is positive since lambda is >1. Is this a good way to start? 2. May 30, 2008 ### Dick Not really. I think you are supposed to use induction. Prove the statement is true for n=1. Then prove is the statement is true for n, then it's true for n+1. Isn't that how induction is supposed to go? What you've done so far doesn't look very 'inductive'. 3. May 30, 2008 ### rock.freak667 Well what I was essentially trying to do was to find an expression for a_n+1-2^g(n) and then add,subtract,multiply and divide various things to $a_n \geq 2^{g(n)}$ and then eventually say that from my expression a_n+1 >= 2^g(n) 4. May 30, 2008 ### Dick Yeah, I know. But the problem says 'use induction'. And that might be the best way to go. I'll warn you, I haven't actually tried to do it. 5. May 30, 2008 ### exk As Dick mentioned induction may be the way to go: [tex]a_2=2^{\lambda}$$

Take your expression now and show that the statement is true for n=2 (I guess this is also your inductive hypothesis):

$$\frac{(a_n^2+1)^{\lambda}}{a_n^{\lambda}}-2^{g(n)} \geq 0 --> \frac{(a_n^2+1)}{a_n} \geq 2^{\lambda^{n-2}}$$

So then:
$$\frac{(a_2^2+1)}{a_2} \geq 2 --> \frac{(2^{2\lambda}+1)}{2^{\lambda}} \geq 2$$

So now you need to find a way to express the relation for n=k+1. I guess expanding a couple of terms may be helpful, but I always came back to the general form that you proposed.

6. May 30, 2008

### rock.freak667

Then just say:

Assume true for n=N

$a_N \geq 2^{g(n)}$

+ $\frac{1}{a_N}$

$$a_N +\frac{1}{a_N} \geq 2^{g(n)} +\frac{1}{a_N}$$

Raise both sides to the power of $\lambda$

$$(a_N +\frac{1}{a_N})^{\lambda} \geq (2^{g(n)} +\frac{1}{a_N})^{\lambda}$$

and then say that

$$(2^{g(n)} +\frac{1}{a_N})^{\lambda} \geq 2^{g(N+1)}$$

and so

$$a_{N+1}>2^{g(N+1)}$$

and hence true by PMI ?

7. May 30, 2008

### Vid

We need to show that

(an + 1/an)^t >= 2^(t^(n))
Observation:
t*ln(an + 1/an) >=t^(n)ln(2)
ln(an + 1/an) >= t^(n-1)(ln(2))
an + 1/an >= 2^(t^(n-1))
Since t >= 1, it's an increasing sequence. Thus,
an + 1/an > an >= 2^(t^(n-1)) by the inductive hypothesis. QED.

8. May 30, 2008

### Vid

I see no problems with your proof.

9. May 31, 2008

### rock.freak667

But I might need to justify why

$$(2^{g(n)} +\frac{1}{a_N})^{\lambda} \geq 2^{g(N+1)}$$

so I would just say that the 2^g(n)+(anything) must be bigger than 2^(g(n+1)) since a_n is greater than zero

10. May 31, 2008

### Vid

(2^(t^(n-1) + 1/an)^t >= 2^(t^(n-1)*t) = 2^(t^n) = 2^(g(n+1)). The inequality holds since t> 1 and an is postive.