Consider a sequence $a_{1}=1$ and for every $k>1$ integer $a_k=a_{k-1}+\dfrac{1}{a_{k-1}}$.
a) How many positive integers $n$ are there, satisfying that $a_n$ be an integer?
b) Find the limit (if there is) of the $a_k$ sequence!
Please help! It looks like that for a) the answer is 2. Thanks in advance!
On
Let's prove that $$ 1 \le a_n \le n $$ Since $a_1 = 1$ and $a_{n+1} > a_n$ we clearly have $a_n \ge 1$. We can prove that $a_n \le n$ by induction. We have $a_1 = 1 \le 1$ and induction step is $$ a_{n+1} = a_n + \frac{1}{a_n} \le n + \frac{1}{a_n} \le n + 1, $$ since $a_n \ge 1$.
About question a). If $a_n$ is an integer then $1/a_n$ is an integer only if $a_n = 1$. That gives us case $a_2 = 2$. Also by definition $a_1 = 1$. There is no other integers in this sequence. Let $a_n = \frac{p}{q}$ where $p$ and $q$ are coprime and $q>1$. Then $$ a_{n+1} = a_n + \frac{1}{a_n} = \frac{p^2 + q^2}{pq}. $$ So we must have $p^2 + q^2 = m\cdot pq$ for some $m \in \mathbb{N}$. But that's not possible if $p$ and $q$ are coprime, because $q$ must divide $p^2$ and $q > 1$. So there are only two indices ($n=1,2$) for which $a_n$ is integer.
Now regarding question b). Since $$ a_{n+1} = a_n + \frac{1}{a_n} \ge \frac{1}{a_n} + \frac{1}{a_{n-1}} + \ldots + \frac{1}{a_1} \ge \frac{1}{n} + \frac{1}{n-1} + \ldots + 1 = H_n $$ where $H_n$ is a harmonic series, which clearly diverges.
At the limit point, $a_k=a_{k-1}$, hence $a_{k-1}=a_{k-1}+\frac{1}{a_{k-1}}$, which implies $\frac{1}{a_{k-1}}=0$, which can't exist, so $a_{k-1}\to \infty$, as does $a_k$.
Let $a_{k-1}=\frac{x}{y}$. Then: $a_k=\frac{x}{y}+\frac{y}{x}=\frac{x^2+y^2}{xy}$.
$$\frac{x^2+y^2}{xy}\in \Bbb Z \iff x=y, \text{where} \frac{x^2+x^2}{x^2}=2$$ Since $x\ne y$ for every successive case (we can infer this because it is an increasing function and tends to $\infty$), your only integer cases are $a_1=1$ and $a_2=2$.