TITLE: Prove that $A_{100} \gt 14$ where $A_{n}=A_{n-1}+\frac{1}{A_{n-1}}$ and $A_1=1$
QUESTION [6 upvotes]: I tried attempting the question, and the best upper bound I could obtain was $1+\ln{98}$. I tried using $A_{n}\le n$ to form a harmonic series, but that wasn't strong enough. Any help would be appreciated, Thanks.

REPLY [2 votes]: Another fleshing out of Michael's answer:
Let $B_n = A_n^2$. Then $B_n=A_{n-1}^2 + 2 + \frac{1}{A_{n-1}^2} > B_{n-1} + 2$.
Thus, $B_n > B_1 + (n-1)2=2n-1$.
Now, $A_n > 14$ iff $B_n > 196$, and so it suffices to take $2n-1>196$, that is, $n \ge 99$.