Suppose {a_{1}} 0 and a_{n+1} = a_{n} + 1/a_{n}. Prove that {a_{n}} diverges to infinity

Question

Suppose {a_{1}} > 0 and a_{n+1} = a_{n} + 1/a_{n}. Prove that {a_{n}} diverges to infinity

anonymous · Answer

Suppose $a_{1}\ > 0$ and $a_{n+1}= a_{n} + \frac{1}{a_{n}}$. Prove $a_{n}\ \rightarrow\ \infty$

zarkon · Answer

here is a hint

either $a_1\ge1$ or $\dfrac{1}{a_1}\ge 1$

anonymous · Answer

Couldnt there be a possibility that $0< a_{n} < 1$?

zarkon · Answer

$$a_2=a_1+\frac{1}{a_1}$$

$$a_3=a_2+\frac{1}{a_2}$$

$$a_4=a_3+\frac{1}{a_3}=a_2+\frac{1}{a_2}+\frac{1}{a_2+\frac{1}{a_2}}>a_2+\frac{1}{a_2}+\frac{1}{a_2+1}$$

$$a_5=a_4+\frac{1}{a_4}>a_2+\frac{1}{a_2}+\frac{1}{a_2+1}+\frac{1}{a_2+\frac{1}{a_2}+\frac{1}{a_2+\frac{1}{a_2}}}$$

$$>a_2+\frac{1}{a_2}+\frac{1}{a_2+1}+\frac{1}{a_2+1+1}=a_2+\frac{1}{a_2}+\frac{1}{a_2+1}+\frac{1}{a_2+2}$$


$$a_n>a_2+\sum_{k=0}^{n-3}\frac{1}{a_2+k}$$

zarkon · Answer

the last sum, as n goes to infinity, is essentially the harmonic sum.  That sum diverges to infinity.

anonymous · Answer

Wow. Is there anything that just tips you off to do that, or is it just experience?

zarkon · Answer

well the 1/a1 reminded me on the harmonic.

I also used my hint...can you see where?

anonymous · Answer

Well, I assume it's what allowed you to claim $$a_{2} + \frac{ 1 }{ a_{2} }+ \frac{ 1 }{ a_{2} + \frac{ 1 }{ a_{2} } } > a_{2} + \frac{ 1 }{ a_{2} }+ \frac{ 1 }{ a_{2} + 1 }$$ But that means we're saying that its $$\frac{ 1 }{ a_{n} } \ge 1$$Does that mean we need a case where we assume 
$$a_{n} \ge 1$$ or does it work no matter what with that inequality?

zarkon · Answer

if $a_1>0$ and either $a_1\ge 1$ or $\dfrac{1}{a_1}\ge1$

then $$a_2=a_1+\frac{1}{a_1}>1$$ and therefore $\dfrac{1}{a_2}<1$

anonymous · Answer

Oh, I think I see what you mean. So as you continue on with your successive terms, you can say $$\frac{ 1 }{ a_{k} } < 1$$ for any k >= 2. At least so it seems. But I see what you did more clearly now. Thank you :)

zarkon · Answer

yes