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- Feb 14, 2012

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I worked with this problem for at least an hour but I don't see how to crack it using my way (which I will show below)...I was wondering if my way of thinking is just plain wrong and I won't be able to get anywhere with it.

Problem:

Let $a_n$ be a sequence of positive real numbers defined by $a_1=1$ and $a_{n+1}=a_n+\dfrac{1}{a_n}$ for $n \ge 1$.

Find $\lfloor a_{100000} \rfloor$.

Attempt:

I tried to solve the problem by relating $a_2, a_3, a_4, a_5, a_6, a_7$ etc to $a_1$, which is 1 and hopefully I could "deduce" some useful pattern from it but to no avail...

$a_1=1$

$a_2=\dfrac{a_1}{1}+\dfrac{1}{a_1}=1+\dfrac{1}{1}=2$

$a_3=a_2+\dfrac{1}{a_2}=a_1+\dfrac{1}{a_1}+\dfrac{1}{\left(a_1+\dfrac{1}{a_1} \right)}=\dfrac{a_1^2+1}{a_1}+\dfrac{a_1}{a_1^2+1}$

$a_4=a_3+\dfrac{1}{a_3}=\dfrac{a_1^2+1}{a_1}+ \dfrac{a_1}{a_1^2+1}+\dfrac{1}{\left(\dfrac{a_1^2+1}{a_1}+\dfrac{a_1}{a_1^2+1} \right)}=\dfrac{(a_1+1)^2+a_1^2}{a_1(a_1^2+1)}+ \dfrac{a_1(a_1^2+1)}{(a_1+1)^2+a_1^2}$

If I am allowed to conclude at this juncture that I find we can relate $a_n$ and its previous term, $a_{n-1}$ in the following manner:

$a_n$ is the sum of two fractions where the second fraction is the reciprocal of the first fraction and the first fraction, its numerator is the sum of the two square terms of the two numerators of the previous term ($a_{n-1})$ and the denominator of it would be the product of the two numerators of that previous term.

But I really don't see how this is going to help because I am incapable of "deducing" a correct general formula for $a_n$ based on my observation and hence I hope someone could help me with this hard (at least for me) problem.

Thanks in advance.