$$x + \frac1x + \cfrac1{x + \frac1x} + \cfrac1{x + \frac1x + \cfrac1{x + \frac1x}} +\cfrac1{x + \frac1x + \cfrac1{x + \frac1x} + \cfrac1{x + \frac1x + \cfrac1{x + \frac1x}} } + \cdots $$
How can this series be simplified? Consider me as a high school graduate.
On
Let $a_n$ be the nth partial sum. The series is divergent. Note that since $a_{n+1}=a_n+\frac{1}{a_n}$, $(a_{n+1})^2=(a_n)^2+2+(a_n)^{-2} >(a_n)^2+2$. Therefore, by induction, $(a_n)^2>2n$ for all $n>2$, thus $a_n>\sqrt{2n}$ which means the series is divergent.
Credits to http://www.math.umd.edu/highschool/mathcomp/2000sol2.html for the solution.
Note: The solution above only works for $x>0$. If $x<0$, notice that the partial sum $\mid a_n \mid$ is the same for $x$ and $-x$, so the series diverges for $x<0$ as well.
By setting $a_1=x$ and $a_N = \frac{1}{a_1+a_2+\ldots+a_{N-1}}$ we may easily see that there are quite a lot convergence issues. If $a_1+a_2+a_N\to L\neq 0$, then $a_N\to \frac{1}{L}$, contradicting the convergence of the series. On the other hand also $a_1+a_2+\ldots+a_N\to 0$ leads to a contradiction, hence your series cannot be convergent.