If $a_1=1; a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$ if $n>1$ then $a_{100}$ = ? Problem : 
If $a_1=1; a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$ if $n>1$ then $a_{100}$ = ? 
Putting the value of n=2,3,4, we get : 
when n =2 ; $a_2=1+\frac{1}{1} = 2$ 
when n=3; $a_3 =2+\frac{1}{2}$
when n=4; $a_4 =\frac{5}{2} +\frac{2}{5}$ 
Now how to get the $a_{100}$ term, please suggest thanks. 
 A: To get a good approximation, note that $a_n^2 = \left(a_{n-1}+\dfrac{1}{a_{n-1}}\right)^2 = a_{n-1}^2+2+\dfrac{1}{a_{n-1}^2}$
Hence, $a_{100}^2 = a_1^2 + \displaystyle\sum_{n = 2}^{100}\left(a_n^2-a_{n-1}^2\right) = 1 + \sum_{n = 2}^{100}\left(2+\dfrac{1}{a_{n-1}^2}\right) = 199 + \sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2}$. 
Since $a_n$ is strictly increasing, $a_n \ge a_3 = \frac{5}{2}$ for $n \ge 3$. 
Thus, $\displaystyle\sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2} = \dfrac{1}{a_1^2}+\dfrac{1}{a_2^2}+\sum_{n = 4}^{100}\dfrac{1}{a_{n-1}^2} \le 1 + \dfrac{1}{4} + 97 \cdot \dfrac{1}{\left(\frac{5}{2}\right)^2} = 16.77$. 
Also, $\displaystyle\sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2} = \dfrac{1}{a_1^2}+\dfrac{1}{a_2^2}+\sum_{n = 4}^{100}\dfrac{1}{a_{n-1}^2} \ge 1 + \dfrac{1}{4} + 97 \cdot 0 = 1.25$.
Therefore, $200.25 \le a_{100}^2 \le 215.77$, and thus, $14.15 \le a_{100} \le 14.69$. 
If you compute $a_n$ for a few more small values of $n$, you can get tighter bounds on $\displaystyle\sum_{n = 2}^{100}\dfrac{1}{a_{n-1}^2}$, and hence, a tighter bounds on $a_{100}$. 
