TITLE: Let us define a series $\{a(n)\}$ such that $a(n)= a(n-1) + \frac{1}{a(n-1)}$
QUESTION [7 upvotes]: Let us define a series $\{a(n)\}$ such that 
$a(n)= a(n-1) + \frac{1}{a(n-1)}$  and $a(1)=1$
then prove that $a(75)$ belongs to the interval $(12,15)$
note here i have used $X(n)$ where n is in  sub script denoting the  series count.
Im unable to think..
what i have done is i have added all the terms as $\{ a(n)-a(1)\}$ which gave me:
$a(n)-a(1)=\frac{1}{a(1)} +\frac{1}{a(2)}+\frac{1}{a(3)}......+\frac{1}{a(n-1)}$
i tried am-gm inequality but it is of no help.
Please Help .

REPLY [9 votes]: Notice first that
$$ a_n^2 = a_{n-1}^2 + 2 + \frac{1}{a_{n-1}^2} > a_{n-1}^2 + 2. \tag{*}$$
Recursively applying this inequality and using the fact that $a_2 = 2$,
$$ \forall n \geq 2, \qquad a_n^2 \geq 2n \tag{1} $$
Now plugging this inequality back to $\text{(*)}$, for $n \geq 3$ we have
$$ a_n^2 \leq a_{n-1}^2 + 2 + \frac{1}{2(n-1)} $$
with strict inequality when $n \geq 3$. Recursively applying this inequality yields
$$ a_n^2
\leq a_2^2 + \sum_{k=2}^{n-1} \left(2 + \frac{1}{2k} \right)
= 2n + \sum_{k=2}^{n-1} \frac{1}{2k} \tag{2} $$
When $n = 75$, we obtain a lower bound from $\text{(1)}$:
$$ 12^2
< 150
\leq a_{75}^2 $$
and an upper bound from $\text{(2)}$:
$$ a_{75}^2
\leq 150 + \sum_{k=2}^{74} \frac{1}{2k}
< 150 + \frac{\log 74}{2}
< 150 + \frac{\log_2 128}{2}
< 169
= 13^2. $$
This proves that $ 12 < a_{75} < 13 $ and hence the claim follows.